Question

A soccer ball is kicked upward from a height of 6 ft with an initial velocity of 96 ft/s. How high will it go? Use -32 ft/s? forthe acceleration caused by gravity. Ignore air resistance.

Answer 1

Given,

The initial height of the soccer ball, h₁=6 ft

The initial velocity of the ball, u=96 ft/s

The acceleration due to gravity, g=-32 ft/s

When the ball reaches the maximum height, its velocity will reduce to zero.

Thus the velocity of the ball when it is at its maximum height is v=0 ft/s

From the equation of motion,

`v^2-u^2=2gh_2`

Where h₂ is the total height covered by the ball from its initial height to reach its maximum height.

On substituting the known values,

`\begin{gathered} 0-96^2=2*-32* h_2 \\ \Rightarrow h_2=(-96^2)/(2*-32) \\ =144\text{ ft} \end{gathered}`

Thus the maximum height reached by the ball is,

`\begin{gathered} H=h_1+h_2 \\ =6+144 \\ =150\text{ ft} \end{gathered}`

Thus the maximum height reached by the ball is 150 ft.