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:

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:

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

Answer 2

The maximum height that the rocket reaches is 36feet.

Given function is:

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

Where t = time in seconds since the rocket was shot.

f(t) = the rocket’s height in feet.

How to find out the maximum height?

In order to get the maximum height, we need to find out the time when the rocket will reach maximum height.

For this, $f'(t) =0$ and find t, again check f''(t), if its value is negative, put t in f(t) to get the maximum height.

$f'(t) = \frac{d}{dt} (-16t^2+48t) =0$

-32t + 48 =0

t = 1.5

f''(t) = -32, This means t=1.5 will be the time when the rocket will be at maximum height, or mathematically t will be the point of local maxima.

So, maximum height = f(1.5) = -16*(1.5)² + 48*1.5 = 36feet

Therefore, the maximum height that the rocket reaches is 36feet.

To get more about maxima and minima visit:

https://brainly.com/question/82347