Question

Emmanuel kicks a foot ball off the ground and in the air with an initial velocity of 42 feet per second. Using the formula H(t)=-16t^2 + vt + s, what is the maximum height the soccerball reaches?

Answer 1

This equation is the classic distance with respect to gravity equation.

The key to solving this one is that you need to consider your given initial conditions and then relate them to your equation. You are given the Emmanuel is kicking from the ground.. therefore your start height (s in your formula) is 0. Also, the initial velocity is 42 feet per second. This is v in the term vt. Substituting,

\[H(t)=-16t^{2}+42t+0\]

To see if our equation is right at time 0, the ball should still be on the ground. Substituting t=0, we find that so far we are okay. What we are trying to do is find the critical points for this equation and we can do that by setting the equation equal to zero and then solving for t.

\[-16t^{2} + 42t = 0\]
\[t(-16t+42) = 0\]

An obvious solution is t=0 (we already knew that!). The other one is completed with some simple algebra:

\[-16t+42 = 0, -16t = -42, t = 42/16, t = 21/8\]

So, what this means is that t = 0, the ball is on the ground ready to be kicked. At t = 21/8 the ball hits the ground. Halfway between these two times, the ball is at its maximum height (t=21/16).

So, substitute t = 21/16 into your original equation for the maximum height.

\[H(21/16)=-16(21/16)^{2}+42(21/16)\]

I come up with something around 27.5 feet. Hope this helps.