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,
Mathematics
katherineromano
Question
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?
CAN ANYONE HELP ME ASAP!!!!
Thank you in advance
CAN ANYONE HELP ME ASAP!!!!
Thank you in advance
2 Answer

1. User Answers kate200468
[tex]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}=1055=100\ [ft]\\\\Ans.\ t_{max}=2.5\ second,\ \ h_{max\ rocket}=100\ feet.[/tex] 
2. User Answers carlosego
For this case we have the following function:
[tex] h (t) =  16t ^ 2 + 80t + 5
[/tex]To find the time when it reaches its maximum height, what we must do is to derive the function.
We have then:
[tex] h '(t) =  32t + 80
[/tex]We set zero and clear the time:
[tex] 32t + 80 = 0
32t = 80
[/tex][tex] t =\frac{80}{32}
t = 2.5 s
[/tex]Then, we evaluate the time obtained for the function of the height.
We have then:
[tex] h (2.5) =  16 * (2.5) ^ 2 + 80 * (2.5) +5
h (2.5) = 105 feet
[/tex]Answer:
It will take the rocket to reach its maximum height:
[tex] t = 2.5 s
[/tex]the maximum height is:
[tex] h (2.5) = 105 feet [/tex]