A model rocket is fired vertically upward from rest.

Question

A model rocket is fired vertically upward from rest. Its acceleration for the first three seconds is a(t) = 60t at which time the fuel is exhausted and it becomes a freely "falling" body. Fourteen seconds later, the rocket's parachute opens, and the (downward) velocity slows linearly to -18 ft/s in 5 s. The rocket then "floats" to the ground at that rate.

(a) At what time does the rocket reach its maximum height? (Give your answer correct to one decimal place.)

(b) What is that height? (Give your answer correct to the nearest whole number.)


(c) At what time does the rocket land? (Give your answer correct to one decimal place.)

Answer 1

Hey Mia.

 

We pretty much have 4 different scenarios that we need to treat separately. The result from each scenario will influence the one right afterward.

 

Scenario 1: acceleration upward.

Scenario 2: free-fall.

Scenario 3: parachute deceleration.

Scenario 4: parachute free-fall.

 

**

 

Scenario 1: acceleration upward.

 

It accelerates at a = 60t for three seconds. We'll assume it starts from the ground, i.e. a height of 0.

 

To find the velocity at the end of the acceleration, take the integral.

 

∫ a(t), [t1,t2] = v(t).

 

∫ 60t = 30*t^2, [0,3] = 30*3^2 = 270 ft/s

 

To find the distance traveled throughout the acceleration, integrate the velocity.

 

∫ 30*t^2 = 10*t^3, [0,3] = 10*3^3 = 270 ft

 

This concludes Scenario 1.

 

**

 

Scenario 2: free-fall.

 

Your rocket becomes a projectile under the influence of gravity, so it will follow this equation:

 

y(t) = -1/2*g*t^2 + v0*t + y0

 

Since you're dealing in ft/s, g will be 32.

From Scenario 1 you learned that v0 and y0 are 270 ft/s and 270 ft, respectively.

 

y(t) = -16*t^2 + 270*t + 270

 

To find the time at which t reaches a maximum, use the following equation from Algebra I

 

tmax = -b/2a

tmax = -270/(2*(-16)) = 270/32 = 8.4375 sec, or to 1 decimal place, 8.4 sec.

 

ymax = y(8.4) = -16*(8.4)^2 + 270*(8.4) + 270 = 1409 ft

 

So your rocket's maximum height is 1,409 feet which occurs 8.4 seconds into Scenario 2. But since Scenario 2 didn't start until Scenario 1 ended, you need to add the time elapsed during Scenario 1 to figure out exactly how long it took for the rocket to reach its height.

 

Since Scenario 1 spanned 3 seconds, add this to your tmax from Scenario 2.

 

t = 11.4 seconds to reach the max height.

 

Scenario 3 doesn't begin until 14 seconds after Scenario 2 starts.

 

To find the height at t = 14, simply plug it into the equation:

 

y(14) = -16*14^2 + 270*14 + 270 = 914 ft

 

To find the velocity at t = 14, recognize that at the max height, velocity was 0. The max height occurred 8.4 seconds into Scenario 2, leaving the rocket with 5.6 seconds of free-fall under gravity.

 

v(14) = g*5.6 = -32*5.6 = -179.2 ft/s

 

This concludes Scenario 2.

 

**

 

Scenario 3: parachute deceleration.

 

The problem states that you decelerate linearly to -18 ft/s in 5 seconds. Since you started at -179.2 ft/s, your deceleration is:

 

a = (-18 - -179.2)/5 = 32.2 ft/s^2

 

This makes sense that your acceleration is positive, since it's slowing down your downward velocity.

 

You're interested in your height at the end of Scenario 3.

 

y(5) = (1/2)*(32.2)*5^2 - 179.2*5 + 914 = 420.5 ft

 

This concludes Scenario 3.

 

**

 

Scenario 4: parachute free-fall.

 

You fall at a constant velocity of -18 ft/ s for the rest of your fall, meaning there's no acceleration to account for. You learned in Scenario 3 that you have 420.5 ft to fall, so it will take

 

420.5 ft/ 18 ft/s = 23.4 sec

 

to reach ground.

 

Now sum up all the time from your Scenarios.

 

3 sec of acceleration

+ 14 sec of free-fall

+ 5 sec of parachute deceleration

+ 23.4 sec of parachute free-fall

 

= 45.4 seconds till it reaches the ground.

 

Hope this helps.