Question

3. A model rocket is launched from the ground with an initial velocity of 352 ft/sec.

Answer 1

Answer:

e. It will take 11 seconds to reach the maximum height of 1,936 feet.

f. It will take 22 seconds to return to the earth.

Step-by-step explanation:

Given:

Initial velocity $v_0$ = 352 ft/sec

Solving for question e.

To find the time required to reach the maximum height we will use the formula,

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

where $v_0$ is the starting velocity

$h_0$ is the initial height.

Using the velocity and starting height from our problem we have,

$h(t) = -16t^2+352t+0$,

The path of this rocket will be a downward facing parabola, so there will be a maximum.

This maximum will be at the vertex of the graph.

To find the vertex we start out with $x= \frac{-b}{2a}$ which in our case is,

$x=\frac{-352}{2(-16)}=\frac{-352}{-32}= 11$

So, It will take 11 seconds for the rocket to reach its maximum height.

We will find maximum height using the formula by substituting value of t we get,

$h(11)=-16(11^2)+352(11)+0\\h(11) = -16 \times121+ 352 \times 11 = -1936+3872= 1936 \ ft$

Hence the maximum height will be $1936 \ ft$

Now Solving for question f.

To find the time required for rocket to reach earth.

We will set our formula to 0 to find the time.

$0= -16t^2+352t+0\\-16t(t-22)=0$

Using the zero product property, we know that either -16t = 0 or t - 22 = 0.  When -16t = 0 is at t = 0, when the rocket is launched. t - 22 = 0 gives us an answer of t = 22.

So the rocket reaches the Earth again at 22 seconds.