Question

A soccer ball is kicked upward from a height of 6 ft with an initial velocity of 64 ft/s. How high will it go? Use - 32 ft/s2 forthe acceleration caused by gravity. Ignore air resistance.

Answer 1

Given,

The initial height of the ball, h₁=6 ft

The initial velocity of the ball, u=64 ft/s

The acceleration due to gravity, g=-32 ft/s²

The velocity of the ball when it reaches the maximum height will be, v=0 ft/s

From the equation of motion,

$v^2-u^2=2gh_2$

Where h₂ is the height covered by the ball to reach the maximum height.

On substituting the known values,

$\begin{gathered} 0-64^2=2\times-32\times h_2 \\ h_2=\frac{-64^2}{2\times-32} \\ =64\text{ ft} \end{gathered}$

Thus the maximum height reached by the ball is

$\begin{gathered} H=h_1+h_2 \\ =6+64 \\ =70\text{ ft} \end{gathered}$

Thus the maximum height reached by the ball is 70 ft