An astronaut on a distant planet wants to determine its acceleration due to gravity. The astronaut throws a rock straight up with a velocity of 19 m/s and measures a time of 24.4 s before the rock returns to his hand. What is the acceleration (magnitude and direction) due to gravity on this planet

Answers

Answer 1

Answer:

1.56 m/s²

Explanation:

Projectile motion is a form of motion where an object moves in parabolic path (trajectory). Projectile motion only occurs when there is one force applied at the beginning on the trajectory, after which the only interference is from gravity.

The total time (time of flight) of an object is given by:

T = 2usinθ / g

where u is the initial velocity, θ is the angle with horizontal and g is the acceleration due to gravity

Since the astronaut throws a rock straight up, hence θ = 90°, u = 19 m/s, T = 24.4 s.

T = 2usinθ / g

Substituting:

24.4 = 2(19)(sin90)/g

g = 2(19)(sin90) / 24.4

g = 1.56 m/s²


Related Questions

A proton is released from rest at the positive plate of a parallel-plate capacitor. It crosses the capacitor and reaches the negative plate with a speed of 54000 m/s. The experiment is repeated with a He+ ion (charge e, mass 4 u).What is the ion's speed at the negative plate?

Answers

4058 m longest serving

why clinical thermometer cannot be used to measure the boiling point of water​

Answers

Answer:

: No, a clinical thermometer cannot be used to measure the temperature of boiling water because it has a small range and might break due to extreme heat. ... The temperature is around 100 degrees Celsius.

g Three masses are located in the x- y plane as follows: a mass of 6 kg is located at (0 m, 0 m), a mass of 4 kg is located at (3 m, 0 m), and a mass of 2 kg is located at (0 m, 3 m). Where is the center of mass of the system

Answers

Answer:

Xcm = (6 * 0 + 4 & 3 + 2 * 0) / 12 = 1

Ycm = (6 * 0 + 4 * 0 + 2 * 3) / 12 = 1/2

(Xcm , Ycm) = (1 , 1/2)

Using definition of center of mass

If earth is compressed to the volume of moon, its acceleration due to gravity
* i. decreases
ii. remains same as before
iii. increases
iv. none of these ​

Answers

Answer:

increase

Explanation:

hope it will help you:)

The mass of the sun is 2*10^30 kg and its radius is
6.96*10^8 m. what is the weight of 1kg mass on the
surface of the sun.

Answers

Explanation:

Distance d=1.5×108 km=1.5×1011 m

Mass of the sun, m=2×1030 kg

Mass of the earth, M=6×1024 kg

Force of gravitation, F=G×d2m×M

F=6.7×10−11×(1.5×1011)22×1030×6×1024=3.57×1022 N

A Catapult throws a payload in a circle with an arm that is 65.0 cm long. At a certain instant, the arm is rotating at 8.0 rad/s and the angular speed is increasing at 40.0 rad/s2. For this instant, find the magnitude of the acceleration of the payload.

Answers

Answer:

The acceleration of the payload is 26 m/s2.

Explanation:

length, L = 65 cm =  0.65 m

angular acceleration = 40 rad/s^2

The acceleration is given by

a = angular acceleration x length

a = 40 x 0.65

a = 26 m/s^2

suppose a car of 1200kg is moving with a velocity of 40km/hr therefore its kinetic energy is not zero. 1. explain briefly what happens to its kinetic energy when the driver applies the breaks and the car stops​

Answers

Answer:

Explanation:

For starters begin with a warning not to touch the brake drums. All of the KE is transferred to the brake drums. The result is a large rise in temperature. Heat. If you press hard on the brakes, rubber is left on the road and there is heat involved in that too.

Answer:

KInetic energy reduces.

Explanation:

Application of breaks reduces velocity. Reduction of velocity constitutes velocity reduction.

For more help contact +1 (304) 223-3136

Topic: Chapter 10: Projectory or trajectile?
Projectile range analysis:
A projectile is launched from the ground at 10 m/s, at
an angle of 15° above the horizontal and lands 5.1 m away.
What other angle could the projectile be launched at, with the same velocity,
and land 5.1 m away?

90°
75°
45
50°
30°

Answers

Answer:

The other angle is 75⁰

Explanation:

Given;

velocity of the projectile, v = 10 m/s

range of the projectile, R = 5.1 m

angle of projection, 15⁰

The range of a projectile is given as;

[tex]R = \frac{u^2sin(2\theta)}{g}[/tex]

To find another angle of projection to give the same range;

[tex]5.1 = \frac{10^2 sin(2\theta)}{9.81} \\\\100sin(2\theta) = 50\\\\sin(2\theta) = 0.5\\\\2\theta = sin^{-1}(0.5)\\\\2\theta = 30^0\\\\\theta = 15^0\\\\since \ the \ angle \ occurs \ in \ \ the \ first \ quadrant,\ the \ equivalent \ angle \\ is \ calculated \ as;\\\\90- \theta = 15^0\\\\\theta = 90 - 15^0\\\\\theta = 75^0[/tex]

Check:

sin(2θ) = sin(2 x 75) = sin(150) = 0.5

sin(2θ) = sin(2 x 15) = sin(30) = 0.5

how do you calculate voltage drop

Answers

Answer:

Multiply current in amperes by the length of the circuit in feet to get ampere-feet. Circuit length is the distance from the point of origin to the load end of the circuit.

Divide by 100.

Multiply by proper voltage drop value in tables. The result is voltage drop.

Explanation:

In an exciting game, a baseball player manages to safely slide into second base. The mass of the baseball player is 88.9 kg and the coefficient of kinetic friction between the ground and the player is 0.53. (a) Find the magnitude of the frictional force in newtons. N (b) It takes the player 1.7 s to come to rest. What was his initial velocity (in m/s)

Answers

Answer:

Look at explanation

Explanation:

a) Kinetic Friction= μmg

μmg=0.53*88.9*9.8=461.75N

b)  -461.75N=ma

a= -5.19m/s^2

v=v0+at

5.19*1.7=v0

v0=8.81m/s^2

(a) The magnitude of the frictional force will be 461.75N

(b)The initial velocity will be 8.81 m/s.

What is kinetic friction?

A force that acts among sliding parts is referred to as kinetic friction. A body moving on the surface is subjected to a force that opposes its progressive motion

The size of the force will be determined by the kinetic friction coefficient between the two materials.

The given data in the problem is;

μ is the coefficient of kinetic friction= 0.53.

m is the mass = 88.9 kg

g is the acceleration due to gravity= 9.81 m/s²

v is the speed =?

The formula for friction force is;

[tex]\rm F= \mu R \\\\ R=mg \\\\ F= \mu mg \\\\\ F=0.53 \times 88.9 \times 9.81 \\\\ F= 461.75 \ N[/tex]

Mechanical force is found as;

F=ma

-461.75=(88.9)a

(-ve shows the -ve work done)

a=-5.19 m/s

From the Newton's first equation of motion;

v=u+at

0=u+at

u=-at

u=(- (-5.19)(1.7)

u=8.81 m/s²

To learn more about the kinetic friction refer to;

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In the late 19th century, great interest was directed toward the study of electrical discharges in gases and the nature of so-called cathode rays. One remarkable series of experiments with cathode rays, conducted by J. J. Thomson around 1897, led to the discovery of the electron.
With the idea that cathode rays were charged particles, Thomson used a cathode-ray tube to measure the ratio of charge to mass, q/m, of these particles, repeating the measurements with different cathode materials and different residual gases in the tube.
Part A
What is the most significant conclusion that Thomson was able to draw from his measurements?
He found a different value of q/m for different cathode materials.
He found the same value of q/m for different cathode materials.
From measurements of q/m he was able to calculate the charge of an electron.
From measurements of q/m he was able to calculate the mass of an electron.
Part B
What is the distance Δy between the two points that you observe? Assume that the plates have length d, and use e and m for the charge and the mass of the electrons, respectively.
Express your answer in terms of e, m, d, v0, L, and E0.
Part C
Now imagine that you place your entire apparatus inside a region of magnetic field of magnitude B0 (Figure 2) . The magnetic field is perpendicular to E⃗ 0 and directed straight into the plane of the figure. You adjust the value of B0 so that no deflection is observed on the screen.
What is the speed v0 of the electrons in this case?
Express your answer in terms of E0 and B0.

Answers

Answer:

a) He found the same value of q/m for different cathode materials.

b)      y = [tex]- \frac{e}{m}\ \frac{E_o v_o^2 }{2d^2}[/tex] ,  c)  v = [tex]\frac{E_o}{B_o}[/tex]

Explanation:

In Thomson's experiments he was able to measure the deflection of the light beam under the effect of the magnetic field and with these results find the e / m relationship, which in all cases is the same, therefore the most important conclusion is that the value e E / m is constant for all materials.

b) In the part of the plates the electrons are accelerated by the electric field,

              F = ma

             - e E = m a

              a = - (e/m)  E₀

               

the distance traveled is          

X axis

          x = v₀ t

the separation of the plates is x = d

          t = vo / d

               

Y axis

          y = v_{oy} t + ½ to t²

          y = ½ a t²

          y = [tex]- \frac{e}{m}\ \frac{E_o v_o^2 }{2d^2}[/tex]

           

c) In this case there is a magnetic field B₀ and the electrons have no deflection

         F = - e E + e v x B

       

if there is no deviation F = 0

         e E = e v B

         v = [tex]\frac{E_o}{B_o}[/tex]

A block of mass 10kg is suspendet at a diameter of 20cm from the centre of a uniform bar im long, what force is required to balance it at its centre of gravity by applying the fore at the other end of the bar?​

Answers

Answer:

4 kg of force

Explanation:

Force = (mass x distance to fulcrum) / length of fulcrum to end

Subsitute values

F = (10 x 20)/50

F =4

A 31 kg block is initially at rest on a horizontal surface. A horizontal force of 83 N is required to set the block in motion. After it is in motion, a horizontal force of 55 N i required to keep it moving with constant speed. From this information, find the coefficients of static and kinetic friction

Answers

Answer:

The static and kinetic coefficients of friction are 0.273 and 0.181, respectively.

Explanation:

By Newton's Laws of Motion and definition of maximum friction force, we derive the following two formulas for the static and kinetic coefficients of friction:

[tex]\mu_{s} = \frac{f_{s}}{m\cdot g}[/tex] (1)

[tex]\mu_{k} = \frac{f_{k}}{m\cdot g}[/tex] (2)

Where:

[tex]\mu_{s}[/tex] - Static coefficient of friction, no unit.

[tex]\mu_{k}[/tex] - Kinetic coefficient of friction, no unit.

[tex]f_{s}[/tex] - Static friction force, in newtons.

[tex]f_{k}[/tex] - Kinetic friction force, in newtons.

[tex]m[/tex] - Mass, in kilograms.

[tex]g[/tex] - Gravitational constant, in meters per square second.

If we know that [tex]f_{s} = 83\,N[/tex], [tex]f_{k} = 55\,N[/tex], [tex]m = 31\,kg[/tex] and [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], then the coefficients of friction are, respectively:

[tex]\mu_{s} = \frac{83\,N}{(31\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}[/tex]

[tex]\mu_{s} = 0.273[/tex]

[tex]\mu_{k} = \frac{55\,N}{(31\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}[/tex]

[tex]\mu_{k} = 0.181[/tex]

The static and kinetic coefficients of friction are 0.273 and 0.181, respectively.

The slope of a d vs t graph represents velocity. Describe 3 ways you know this to be true.

Answers

Answer:

Look at explanation

Explanation:

I only know 1 way, there is another way you can rephrase this using derivatives but that's pretty much the same thing.

The slope is calculated by Δy/Δx so the slope of distance vs time graph is Δd/Δt which is the velocity

what will be the gravitational force between two heavenly bodies if the masses of both are tripled keeping the distance between them constant

Answers

Answer:

If the mass of one of the objects is tripled, then the force of gravity between them is tripled. ... Since gravitational force is inversely proportional to the square of the separation distance between the two interacting objects, more separation distance will result in weaker gravitational forces

A flat, 75-turn, coil is oriented with its plane perpendicular to a uniform magnetic field that varies steadily from 0.00 To 1.20 T in 20.0 ms. The diameter of each coil is 10 cm. Calculate the emf induced in the coil during this time, in volts.

Answers

40 m long what is 6 m on

State whether plastic is biodegradable or non-biodegradable ? Give reasons for your answer.​

Answers

Answer:

non biodegradable

Explanation:

It is non biodegradable because plastic cannot dispose off easily ..

What is the strength of the magnetic field a distance 4.4 mm above the center of a circular loop of radius 0.8 mm and current 474.1 A

Answers

Answer:

B = 0.118 T

Explanation:

From Biot-Savart Law:

[tex]B = \frac{\mu_o I}{2\pi r}[/tex]

where,

B = strength of magnetic field = ?

μ₀ = 4π x 10⁻⁷ Tm/A

I = current enclosed = 474.1 A

r = radius = 0.8 mm = 8 x 10⁻⁴ m

Therefore,

[tex]B = \frac{(4\pi\ x\ 10^{-7}\ Tm/A)(474.1\ A)}{2\pi(8\ x\ 10^{-4}\ m)}[/tex]

B = 0.118 T

3. Calculate the force it would take to accelerate a 50 ka bike at a rate of 3 m/s2 (6 points)

Answers

Answer:

150 N

Explanation:

Given that,

Acceleration (a) = 3 m/s²Mass of the bike (m) = 50 kg

We are asked to calculate force required.

[tex]\longrightarrow[/tex] F = ma

[tex]\longrightarrow[/tex] F = (50 × 3) N

[tex]\longrightarrow[/tex] F = 150 N

A uniform 1500-kg beam, 20.0 m long, supports a 15,000-kg printing press
5.0 m from the right support column (Figure slide 8). Calculate the force
on each of the vertical support columns.

Answers

Answer:

[tex]\mathbf{F_1=4.41*10^4\ N}[/tex]

[tex]\mathbf{F_2 = 1.176*10^5 \ N}[/tex]

Explanation:

The missing image of the figure slide is attached in below.

However, from the model, it is obvious that it is in equilibrium.

As a result, the relation of the force and the torque is said to be zero.

i.e.

[tex]\sum F = 0[/tex] and [tex]\sum \tau = 0[/tex]

From the image, expressing the forces through the y-axis, we have:

[tex]F_1+F_2 = W_B + W_P \\ \\ \implies 9.8(1500+15000) \\ \\ \implies \mathtt{1.617\times 10^5 \ N}[/tex]

Also, let the force [tex]F_1[/tex] be the pivot and computing the torque to determine [tex]F_2[/tex]:

Then:

[tex]F_1(0)+F_2(20.0) = 10.0W_B + 15.0W_P[/tex]

[tex]F_2 = \dfrac{((10*1500)+(15*15000))*9.8}{20.0}[/tex]

[tex]F_2 = 117600 \ N[/tex]

[tex]\mathbf{F_2 = 1.176*10^5 \ N}[/tex]

For the force equation:

[tex]F_1+F_2=1.617*10^5 \ N;[/tex]

where:

[tex]F_2 = 1.176*10^5 \ N[/tex]

Then:

[tex]F_1+1.176*10^5 \ N=1.617*10^5 \ N[/tex]

[tex]F_1=1.617*10^5 \ N-1.176*10^5 \ N[/tex]

[tex]F_1=44100\ N[/tex]

[tex]\mathbf{F_1=4.41*10^4\ N}[/tex]

suppose the tank is open to the atmosphere instead of being closed. how does the pressure vary along

Answers

Answer:

Pressure is more in the open container than the closed one.

Explanation:

The pressure due to the fluid at a depth is given by

Pressure = depth x density of fluid x gravity

So, when the container is open, the atmospheric pressure is also add  up but when  the container is closed only the pressure due to the fluid is there.

So, when the container is open, the pressure is atmospheric pressure + pressure due to the fluid.

hen the container is closed only the pressure due to the fluid is there.

An audience of 2250 fills a concert hall of volume 32000 m^3. If there were no ventilation, by how much would the temperature of the air rise over a period of 2.0 h due to the metabolism of the people (70 W/person)?

Answers

246 and 64 minutes later

A small object A, electrically charged, creates an electric field. At a point P located 0.250 m directly north of A, the field has a value of 40.0 N/C directed to the south. If a second object B with the same charge as A is placed at 0.250m south of A (so that objects A and B and point P follow a straight line), what is the magnitude of the total electric field produced by the two objects at P?

Answers

Answer:

E_total = - 50 N / A

Explanation:

The electric field is a vector magnitude whereby

          E_total = Eₐ + E_b

where the bold letters indicate vectors, in this case the charges of the two objects A and B are the same and they are on the same line

         E_total = - E_a - E_b

         

the electric field for a point charge is

        E_a = [tex]k \ \frac{q_a}{r_a^2 }[/tex]

        qₐ= Eₐ rₐ² / k

indicates that Eₐ = 40.0 N / C

        qₐ = 40.0 0.250²/9 10⁹

        qₐ = 2.777 10⁻¹⁰ C

indicates that the charge of the two points is the same

        qₐ = q_b

      E_total = - k qₐ / rₐ² - k qₐ / (2 rₐ)²

      E_total = [tex]-k \ \frac{q_a}{r_a^2} \ ( 1 + \frac{1}{4} )[/tex]

       

we calculate

       E_total = - 40.0 (5/4)

       E_total = - 50 N / A

Mary needs to row her boat across a 160 m-wide river that is flowing to the east at a speed of 1.5 m/s. Mary can row with a speed of 3.6 m/s. If Mary points her boat due north, how far from her intended landing spot will she be when she reaches the opposite shore? What is her speed with respect to the shore?

Answers

Answer: 66.67 m, 44.44 s

Explanation:

Given

Velocity of flow is [tex]u=1.5\ m/s[/tex]

Mary can row with speed [tex]v=3.6\ m/s[/tex]

Width of the river [tex]y=160\ m[/tex]

Flow will drift the Mary towards east, while Mary boat will cause it to travel in North direction

time taken to cross river

[tex]\Rightarrow t=\dfrac{160}{3.6}\\\\\Rightarrow t=\dfrac{400}{9}\ s[/tex]

Flow will drift Mary by

[tex]\Rightarrow x=ut\\\\\Rightarrow x=1.5\times \dfrac{400}{9}\\\\\Rightarrow x=66.67\ m[/tex]

Velocity w.r.t shore is

[tex]\Rightarrow v_{net}=\sqrt{3.6^2+1.5^2}\\\Rightarrow v_{net}=\sqrt{15.21}\\\Rightarrow v_{net}=3.9\ m/s[/tex]

Question 9 of 10
According to the law of conservation of momentum, the total initial
momentum equals the total final momentum in a(n)
A. Interacting system
B. System interacting with one other system
C. Isolated system
D. System of balanced forces

Answers

Answer:

The answer is C. Isolated System

Answer:

C. Isolated system

Explanation :

∵According to law of  conservation of momentum ,In an isolated system ,the total momentum remains conserved.

A horizontal force of P=100 N is just sufficient to hold the crate from sliding down the plane, and a horizontal force of P=350 N is required to just push the crate up the plane. Determine the coefficient of static friction between the plane and the crate, and find the mass of the crate.

Answers

"down/up the plane" suggests an inclined plane, but no angle is given so I'll call it θ for the time being.

The free body diagram for the crate in either scenario is the same, except for the direction in which static friction is exerted on the crate. With the P = 100 N force holding up the crate, static friction points up the incline and keeps the crate from sliding downward. When P = 350 N, the crate is pushed upward, so static friction points down. (see attached FBDs)

Using Newton's second law, we set up the following equations.

• p = 100 N

F (parallel) = f + p cos(θ) - mg sin(θ) = 0

F (perpendicular) = n - p sin(θ) - mg cos(θ) = 0

P = 350 N

F (parallel) = P cos(θ) - F - mg sin(θ) = 0

F (perpendicular) = N - P sin(θ) - mg cos(θ) = 0

(where n and N are the magnitudes of the normal force in the respective scenarios; ditto for f and F which denote static friction, so that f = µn and F = µN, with µ = coefficient of static friction)

Solve for n and N :

n = p sin(θ) + mg cos(θ)

N = P sin(θ) - mg cos(θ)

Substitute these into the corresponding equations containing µ, and solve for µ :

µ = (mg sin(θ) - p cos(θ)) / (mg cos(θ) + p sin(θ))

µ = (P cos(θ) - mg sin(θ)) / (P sin(θ) + mg cos(θ))

Next, you would set these equal and solve for m :

(mg sin(θ) - p cos(θ)) / (mg cos(θ) + p sin(θ)) = (P cos(θ) - mg sin(θ)) / (P sin(θ) + mg cos(θ))

...

Once you find m, you back-substitute and solve for µ, but as you might expect the result will be pretty complicated. If you take a simple angle like θ = 30°, you would end up with

m ≈ 36.5 kg

µ ≈ 0.256

The coefficient of static friction between the plane and the crate is μ = 0.256 and the mass of the crate is m=36.4 kg.

From the given,

The force that opposes the crate by sliding is P = 100N

In X-axis, the sum of forces is zero.

ΣF = 0

Pcosθ - mgsinθ-Ff = 0

Ff = Pcosθ - mgsinθ

In Y-axis

Psinθ - mgcosθ - N = 0

N = Psinθ-mgcosθ

Frictional force, Ff = μN, μ is the coefficient of friction

Ff = μN

Pcos30- mgsin30 + μ( Psin30+mgcos30) = 0

μ = mgsin30-Pcos30/Psin30+mgcos30 ------1

The block is sliding with the horizontal force, F = 350N

X-axis

P₂cosθ - mgsinθ-Ff = 0

Y-axis

P₂sinθ - mgcosθ - N = 0

N = P₂sinθ-mgcosθ

μ = P₂cos30-mgsin30/P₂sin30-mgcos30   -----2

Equate equations 1 and 2

mgsin30-Pcos30/Psin30+mgcos30 =P₂cos30-mgsin30/P₂sin30-mgcos30

4.905m-86.6/50+8.49 = 303.1-4.905m/175+8.49

41.7m² + 123m - 1.516×10⁴ = 0

-41.7m² +2330m -1.516×10⁴(4.905-86.6)(175+8.49) =(303.1-4.905)(50+8.49)

83.4m² - 2207m -3.03×10⁴ = 0

m= 36.4 kg

Hence, the mass of the crate is 36.4 Kg.

Substitute the value of m in equation 1,

μ = 4.905(36.4) - 86.6 / 50 + 8.49

μ  = 0.256

Thus, the coefficient of static friction is 0.256.

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What are the differences among elements, compounds, and mixtures?

Answers

Answer:

Elements have a characteristic number of electrons and protons.Both Hydrogen(H) and oxygen(O) are two different elements.

••••••••••••••••

Compounds are chemical substances where the atoms atoms of two different elements are combined together. It is made of .Hydrogen(H) and Oxygen(O) both qr4 naturally gases,but they react to form water(H2O),which is liquid compound.

•••••••••••••••

A mixture is made of atleast two parts》 solid,liquid or gas.The difference is that it's not a chemical substance that's bonded by other elements. 

------------------------------

Hope it helps...

Have a great day!!!

Answer: Elements have a characteristic number of electrons and protons. Both Hydrogen(H) and oxygen(O) are two different elements. Compounds are chemical substances where the atoms atoms of two different elements are combined together. It is made of.Hydrogen(H) and Oxygen(O) both qr4 naturally gases, but they react to form water(H2O), which is liquid compound. A mixture is made of at least two parts solid, liquid, or gas. The difference is that it's not a chemical substance that's bonded by other elements.

Think about a thermos bottle. It consists of an inner bottle with a shiny silver surface separated from an outer container by a space with no air. In what ways does it block conduction, convection, and radiation?

Answers

Answer:

Radiation

Explanation:

Conduction, convection and radiation are the three modes of heat transfer.

1. Conduction: When the one end is heated of any rod, the heat transfer to the other end by the vibrational motion of the molecules, it is called conduction.

The heat transfer in a solid is due to the conduction.

2. Convection: When the liquid or gas is heated, the molecules which is in contact to the heat, heated first and due to the decrease in density they moves up and the molecules on the upper side are higher in density so they moves down. These are called convection currents. The process continues till the entire liquid becomes heated. It generally takes place in liquids and gases.  

3. Radiation: The process of heat transfer in which no molecules takes place is called radiation. The heat coming from sun is due to the radiation. It does not require any medium.  

In the thermos bottle, as there is no air between the two layers, so the heat transfer is due to the radiation.  

In 2009 Usain Bolt set the world record time by running 100 meters in 9.58 s. Assume that during this race he ran in a straight line with constant acceleration a. What would be the required constant acceleration a

Answers

Usain Bolt I think, sorry if it’s not right

In 2009 Usain Bolt set the world record time by running 100 meters in 9.58 seconds, assuming that he ran this race with constant acceleration, then the required constant acceleration would have been

What are the three equations of motion?

There are three equations of motion given by  Newton

v = u + at

S = ut + 1/2×a×t²

v² - u² = 2×a×s

By using the second equation of motion given by Newton,

S = ut + 1/2at²

100= 0 + 0.5*a*9.58²

a = 2.17 meters / second²

Thus,the required constant acceleration of Usain Bolt would have been 2.17 meters / second².

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Consider a tall building of height 200.0 m. A stone A is dropped from the top (from the cornice of the building). One second later another stone B is thrown vertically up from the point on the ground just below the point from where stone A is dropped.Birthstones meet at half the height of the tower. (a) Find the initial velocity of vertical throw of stone B.(b) Find the velocities of A and B, just before they meet.

Answers

Answer:

a)  v₀ = 44.27 m / s, b) stone A  v = 44.276 m / s,  stone B   v = 0.006 m / s

Explanation:

a) This is a kinematics exercise, let's start by finding the time it takes for stone A to reach half the height of the building y = 100 m

          y = y₀ + v₀ t - ½ gt²

as the stone is released its initial velocity is zero

         y- y₀ = 0 - ½ g t²

         t = [tex]\sqrt{ -2(y-y_o)/g}[/tex]

         t = [tex]\sqrt{ -2(100-200)/9.8}[/tex]

         t = 4.518 s

now we can find the initial velocity of stone B to reach this height at the same time

         y = y₀ + v₀ t - ½ g t²

stone B leaves the floor so its initial height is zero

         100 = 0 + v₀ 4.518 - ½ 9.8 4.518²

         100 = 4.518 v₀ - 100.02

         v₀ = [tex]\frac{100-100.02}{4.518}[/tex]

         v₀ = 44.27 m / s

   

b) the speed of the two stones at the meeting point

stone A

          v = v₀ - gt

          v = 0 - 9.8 4.518

          v = 44.276 m / s

stone B

          v = v₀ -g t

          v = 44.27 - 9.8 4.518

          v = 0.006 m / s

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