Question

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?

Answer 1

Answer:

a) $v_{0y} = 9.72 m/s$

b) $v_{0x} = 20.85 m/s$

c) $t = 2.01 s$

d) $h_{max} = 4.82 m$

Explanation:

a) The initial vertical velocity is given by:

$v_{0y} = v*sin(\theta)$

Where:

θ: 25°

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

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

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

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

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

$v_{0x} = \frac{x}{t}$

Where:

x: is the total distance = 42 m

$t = \frac{x}{v_{0x}} = \frac{42 m}{20.85 m/s} = 2.01 s$

d) The maximum height is given by:

$v_{fy}^{2} = v_{0y}^{2} - 2gh_{max}$

Where:

$v_{fy}$: is the final vertical velocity =0 (at the maximum heigth)

g: is the gravity = 9.81 m/s²

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

I hope it helps you!