Question

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

e. How long will it take the rocket to reach its maximum height? Show all work in the space provided.












f. Assume the model rocket’s parachute failed to deploy and the rocket fell back to the ground. How long would it take the rocket to return to Earth from the time it was launched? Show all work in the space provided.

Answer 1

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.

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= (-b)/(2a)` which in our case is,

`x=(-352)/(2(-16))=(-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 *121+ 352 * 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.