Question

The height of a toy rocket that is shot in the air with an upward velocity of 48 feet per second can be modeled by the function f(t) = -16t^2 + 48t, where t is the time in seconds since the rocket was shot and f(t) is the rocket’s height in feet. What is the maximum height the rocket reaches?
A) 16 ft
B) 36 ft
C) 48 ft
D) 144 ft

Answer 1

Answer:

36 ft

edg.2020

Step-by-step explanation:

Answer 2

Answer:

B) 36 ft

Step-by-step explanation:

Given:

The height as a function of time is given as:

$f(t)=-16t^2+48t$

At maximum height, the instantaneous velocity becomes 0. The instantaneous velocity is the first derivative of the height function.

So, the derivative of the the given function is 0 at the maximum height.

Differentiating the above function with time, we get

$f'(t)=\frac{d}{dt}(-16t^2)+\frac{d}{dt}(48t)\\f'(t)=-32t+48$

Now, equating the derivative to 0 and finding time.

$f'(t)=0\\-32t+48=0\\32t=48\\t=\frac{48}{32}=1.5\ s$

Therefore, time taken to reach maximum height is 1.5 s.

Now, maximum height is obtained by plugging in $t=1.5$ in the height equation.

Maximum height is given as:

$h_{max}=-16(1.5)^2+48(1.5)\\h_{max}=-16\times 2.25+72\\h_{max}=-36+72\\h_{max}=36\ ft$