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?

Respuesta :

SOLUTION:

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

[tex]\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}[/tex][tex]\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}[/tex]

CONCLUSION:

The maximum height the ball reached was 149 feet.