Question

A model rocket is launched from the ground with an initial velocity of 120 feet per second. The height of an object h, in feet, after t seconds, with initial velocity v0 and initial height h0 is given by h(t)=−16t2+v0t+h0 . What is the approximate maximum height the rocket reaches?

Answer 1

Answer:

The rocket will reach a maximum height of 225 feet from the ground in 3.75 seconds.      

Step-by-step explanation:

We are given the following information in the question:

Initial velocity = $v_0$ = 120 feet per second

Initial heigth = $h_0$

$h(t) = -16t^2 + v_0t + h_0$

h(t) is a function of t that gives height of the rocket at time t, initial velocity $v_0$. intial height $h_0$.

Differentiating h(t) with respect to t, we get:

$\displaysttyle\frac{d(h(t))}{dt} = -32t + v_0$

Equating the first derivative to zero,

$-32t + v_0 = 0\\-32t = -120 \\t = \displaystyle\frac{120}{32} = 3.75$

Again differentiating, h(t) with respect to t,

$\displaysttyle\frac{d^2(h(t))}{dt^2} = -32 < 0$

Hence, h(t) will have a local maxima by double derivative test.

Maximum height attained by rocket =

$h(\displaystyle\frac{120}{32}) = -16\times \frac{120}{32}\times \frac{120}{32} + 120\times \frac{120}{32} + 0 = 225 \text{ feet}$

Hence, the rocket will reach a maximum height of 225 feet from the ground in 3.75 seconds.

Answer 2

567 is the height it reaches