Question

A ball is thrown from ground level with an initial speed of 24.5 m/s at an angle of 35.5 degrees above the horizontal. The ball hits a wall that is 25.8 meters horizontally from where it started. How high (meters) does the ball hit on the wall?

Answer 1

6.07 m

Explanation:

Given:

$v_0=24.5\:\text{m/s}$

$\theta_0 = 35.5°$

First, we need to find the amount of time it takes to travel a horizontal distance of 25.8 m. We know that

$x = v_{0x}t \Rightarrow t = \dfrac{x}{v_0 \cos \theta_0}$

or

$t = 1.29\:\text{s}$

To find the vertical height where the ball hit the wall, we use

$y = v_{0y}t - \frac{1}{2}gt^2$

$\:\:\:\:=(24.5\:\text{m/s})\sin 35.5(1.29\:\text{s}) \\ - \frac{1}{2}(9.8\:\text{m/s}^2)(1.29\:\text{s})^2$

$\:\:\:\:=6.07\:\text{m}$