Question

My teacher gave me the problem : David is throwing in the air at an angle from the ground with an initial velocity of 30 feet per second. The equation h = -10t^2 + 40t + 2 gives the height of the stone thrown. Find the maximum height of the stone.

Answer 1

Given data:

The expression for the height is h=-10t^2 +40t+2.

Equate the derivative of the height to zero in order to mmaximize the height.

$\begin{gathered} \frac{dh}{dt}=0 \\ \frac{d}{dt}(-10t^2+40t+2)=0 \\ -20t+40=0 \\ -20t=-40 \\ t=2 \end{gathered}$

Substitute 2 for t in the given expression of height.

$\begin{gathered} h=-10(2)^2+40(2)+2 \\ =-40+80+2 \\ =42\text{ ft} \end{gathered}$

Thus, the maximum height of the stone is 42 feet.