Question

A soccer ball is kicked horizontally off a 22 m high hill and lands a distance R= 53 m from the edge of the hill. Determine the initial horizontal velocity of the soccer ball.

Answer 1

25.3 m/s

Step-by-step explanation:

First, we need to calculate the time that it takes to reach the ground, so we will use the following equation

`y=v_(iy)t+(1)/(2)gt^2`

Where Viy is the initial vertical velocity, so Viy = 0 m/s, y is the height of 22m and g is the gravity, so g = 10 m/s². Replacing the values and solving for t, we get

`\begin{gathered} 22=0t+(1)/(2)(10)t^2 \\ 22=5t^2 \\ (22)/(5)=t^2 \\ \\ 4.4=t^2 \\ √(4.4)=t \\ 2.1\text{ s = t} \end{gathered}`

Then, with the time t = 2.12 s and the horizontal distance R = 53 m, we can calculate the initial horizontal velocity as

`\begin{gathered} v=(distance)/(time) \\ \\ v=\frac{53\text{ m}}{2.1\text{ s}} \\ \\ v=25.26\text{ m/s} \end{gathered}`

Therefore, the answer is 25.3 m/s