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

Step-by-step 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 = (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 - (1)/(2)gt^2`

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

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