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

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:

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

** Answer: **

** It will take the rocket to reach its maximum height: **

\( t = 2.5 s \)

** the maximum height is: **

\( h (2.5) = 105 feet \)