A projectile was fired from ground with 50 m/s initial velocity at 45 degree angle. How long will stay in the air? What is range? What is the maximum height the projectile?

Answer:

1. Time of flight = 7.22 s

2. Range = 255.10 m

3. Maximum height = 63.77 m

Explanation:

The following data were obtained from the question:

Initial velocity (u) = 50 m/s

Angle of projection (θ) = 45°

Time of flight (T) =?

Range (R) =?

Maximum height (H) =?

1. Determination of the time of flight

Initial velocity (u) = 50 m/s

Angle of projection (θ) = 45°

Acceleration due to gravity (g) = 9.8 m/s²

Time of flight (T) =?

T = 2uSine θ /g

T = 2 × 50 Sine 45 / 9.8

T = 100 × 0.7071 /9.8

T = 7.22 s

2. Determination of the range.

Initial velocity (u) = 50 m/s

Angle of projection (θ) = 45°

Acceleration due to gravity (g) = 9.8 m/s²

Range (R) =?

R = u²Sine 2θ /g

R = 50² × Sine (2×45) /9.8

R = 2500 × Sine 90 /9.8

R = 2500 × 1 / 9.8

R = 255.10 m

3. Determination of the maximum height.

Initial velocity (u) = 50 m/s

Angle of projection (θ) = 45°

Acceleration due to gravity (g) = 9.8 m/s²

Maximum height (H) =

H = u²Sine²θ /2g

H = 50² × (Sine 45)² / 2 × 9.8

H = 2500 × (0.7071)² / 19.6

H = 63.77 m

Answer:

Explanation:

Given;

initial velocity of projection, u = 50 m/s

angle of projection, θ = 45°

Time spent in the air is the time of flight, given as;

[tex]T = \frac{2usin \theta}{g}\\\\T = \frac{2(50)sin (45)}{9.8}\\\\T = 7.22 \ s[/tex]

The range of the projectile is given by;

[tex]R = \frac{u^2sin(2\theta)}{g}\\\\R = \frac{u^2sin(2*45)}{g}\\\\ R = \frac{u^2sin(90)}{g}\\\\ R = \frac{u^2(1)}{g}\\\\R=R_{max} = \frac{u^2}{g} =\frac{50^2}{9.8} = 255.1 \ m[/tex]

The maximum height of the projectile is given by;

[tex]H = \frac{u^2sin^2\theta}{2g}\\\\H = \frac{(50)^2sin^2(45)}{2*9.8}\\\\H = 63.77 \ m[/tex]

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