A toy rocket is shot vertically into the air from a launching pad 5 feel above the ground with an initial velocity of 80 feet per second. The height h, in feet, of the rocket above the ground as t seconds after launch is given by the function h(t)=-16t^2+80t+5. How long will it take the rocket to reach its maximum height? What is the maximum height?

$$h(t)=-16t^2+80t+5\\\\t_{max}-time\ for\ a\ maximum\ height\\\\t_{max}=- \frac{80}{2\cdot(-16)} = \frac{80}{32} =2.5\ [s]\\\\h_{max}-the\ maximum\ height\ above\ the\ ground\\\\h_{max}=h(2.5)=-16\cdot2.5^2+80\cdot2.5+5=-16\cdot6.25+200+5=\\.\ \ \ \ \ \ =-100+205=105\\\\h_{max\ rocket}-the\ maximum\ height\ of\ a\ toy\ rocket\\\\h_{max\ rocket}=105-5=100\ [ft]\\\\Ans. \ t_{max}=2.5\ second,\ \ h_{max\ rocket}=100\ feet.$$

For this case we have the following function:

$$h (t) = - 16t ^ 2 + 80t + 5$$

To find the time when it reaches its maximum height, what we must do is to derive the function.

We have then:

$$h ’(t) = - 32t + 80$$

We set zero and clear the time:

$$-32t + 80 = 0\\32t = 80$$

$$t =\frac{80}{32}\\t = 2.5 s$$

Then, we evaluate the time obtained for the function of the height.

We have then:

$$h (2.5) = - 16 * (2.5) ^ 2 + 80 * (2.5) +5\\h (2.5) = 105 feet$$

$$t = 2.5 s$$
$$h (2.5) = 105 feet$$