122k views
4 votes
If a football is kicked straight up with an initial velocity of 96 ft/sec from a height of 5 ft, then it’s height above the earth is a function of time given by h(t) = -16t^2+96t+5What is the maximum height reached by the ball?

User Jlang
by
6.7k points

1 Answer

4 votes

SOLUTION:

We are to find the maximum height reached by the ball;


\begin{gathered} h(t)=-16t^2+96t\text{ +5} \\ h\text{ ' (t) =}-32t\text{ + 96} \\ -32t\text{ + 96 = 0 } \\ 96\text{ = 32t} \\ t\text{ = 3 seconds} \end{gathered}
\begin{gathered} h(3)=-16t^2+96t\text{ + 5} \\ h(3)=-16(3)^2\text{ + 96(3) + 5} \\ h(3)\text{ = -16(9) + 288+5} \\ h(3)\text{ = -144 + 288 + 5} \\ h(3)\text{ = 149} \end{gathered}

CONCLUSION:

The maximum height the ball reached was 149 feet.

User Sok Chanty
by
6.6k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.