A baseball is thrown at an angle of 25 degrees relative to the ground at a speed of 23 m/s. If the ball was caught 42 m from the thrower:

What is the initial vertical velocity?

What is the initial horizontal velocity?

How long was it in the air?

How high did the ball travel before being caught?

Answers

Answer 1

Answer:

a) [tex] v_{0y} = 9.72 m/s [/tex]

b) [tex] v_{0x} = 20.85 m/s [/tex]

c) [tex] t = 2.01 s [/tex]

d) [tex] h_{max} = 4.82 m [/tex]

Explanation:

a) The initial vertical velocity is given by:

[tex] v_{0y} = v*sin(\theta) [/tex]

Where:

θ: 25°

v: is the magnitude of the speed = 23 m/s

[tex] v_{0y} = 23 m/s*sin(25) = 9.72 m/s [/tex]

b) The initial horizontal velocity can be calculated as follows:

[tex] v_{0x} = v*cos(\theta) = 23 m/s*cos(25) = 20.85 m/s [/tex]

c) The flight time can be calculated using the following equation:

[tex] v_{0x} = \frac{x}{t} [/tex]

Where:

x: is the total distance = 42 m

[tex] t = \frac{x}{v_{0x}} = \frac{42 m}{20.85 m/s} = 2.01 s [/tex]

d) The maximum height is given by:

[tex] v_{fy}^{2} = v_{0y}^{2} - 2gh_{max} [/tex]

Where:

[tex] v_{fy}[/tex]: is the final vertical velocity =0 (at the maximum heigth)

g: is the gravity = 9.81 m/s²

[tex]h_{max} = \frac{v_{0y}^{2}}{2g} = \frac{(9.72 m/s)^{2}}{2*9.81 m/s^{2}} = 4.82 m[/tex]

I hope it helps you!                                  


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