Amodel rocket is launched from a raised platform at a speed of 144 feet per second. its height in feet is given by h( t) = -16 t 2 + 144 t + 20 ( t = seconds after launch). after how many seconds does the object reach its maximum height?

after 4.5 seconds

Step-by-step explanation:

Given the eqn of the height reached by the rocket.

h(t)= -16t^2 +144t +20

Note: at the maximum height dh(t)/dt = 0

Therefore, differentiating the eqn of height.

dh(t)/dt = -32t + 144

Equating dh(t)/dt to zero, we have.

-32t+144=0

32t=144

t = 144/32

t = 4.5 seconds