A rocket is launched upward so that its distance, in feet, above the ground after t seconds is represented by the function below. What is its maximum height? h(t)=−16t2+320t

Given : A rocket is launched upward so that its distance, in feet, above the ground after t seconds is represented by the function [tex]h(t)=-16t^2+320t[/tex]

To find : its maximum height.

First we differentiate the given function , we get

[tex]h'(t)=(2)(-16)t+320=-32t+320[/tex]

Put [tex]h'(t)=0[/tex], we get

[tex]-32t+320=0\\\\\Rightarrow\ t=\dfrac{-320}{-32}=10[/tex]

Hence, at t=10 , rocket achieves its maximum height.

[tex]\text{Maximum Height =}h(t)=-16\left(10\right)^{2}+320\left(10\right)\\\\=-1600+3200=1600[/tex]

Hence, its maximum height = 1600 feet.

lublana lublana · Answer 2 · 2019-10-14T11:41:03+02:00

Answer:

1600 ft

Step-by-step explanation:

We are given that a rocket is a launched upward so that its distance(feet) above the ground after t seconds is represented by the function

[tex]h(t)=-16t^2+320t[/tex]

We have to find the maximum height.

Substitute h(t)=0

[tex]-16t(t-20)=0[/tex]

[tex]t=0,t-20=0\implies t=20[/tex]

When t=0 it means the rocket is at ground launch.

When t= 20 s.

Total time taken by rocket=20 s.

Half of the time taken to reach maximum height and half of the time taken to reach ground back.

Therefore, time taken by rocket to reach maximum height=[tex]\frac{20}{2}=10s[/tex]

Substitute t=10 in given function

Then we get

h(10)=-16(10)^2+320(10)=-1600+3200=1600 ft[/tex]

Hence, the maximum height=1600 ft