Sunday

January 25, 2015

January 25, 2015

Posted by **Michael** on Sunday, February 24, 2008 at 4:55pm.

--------( 20e^(-t/2), for 0 <= t <= 10 seconds)

a(t) = (

--------( -16, for t > 10 seconds)

a) At what time does the rocket begin to descend?

b) How high does the rocket reach?

c) What is the velocity when the rocket impacts the earth?

d) Write a formula for the position of the rocket with respect to time for t > 10 seconds.

Thanks!

- Calc -
**Damon**, Sunday, February 24, 2008 at 6:21pmI believe both drwls and I did this problem not long ago.

It is kind of wacky.

The acceleration should increase, not decrease, while the engine is on for the first ten seconds because the mass decreases as fuel is used up.

The acceleration down during the ballistic phase with the engine off should be -32 ft/s^2, gravity, not -16.

Never the less I will get you started.

a = d^2h/dt^2 = 20 e^-t/2

v = dh/dt = c - 40 e^-(t/2)

since v = 0 at t = 0, c = 40 and

v = 40 (1-e^-(t/2))

Now we can get v at t = 10

v = 39.7 ft/s at t = 10

how high is it at t = 10?

h = c2 + 40 t + 80 e^(-t/2)

h = 0 when t = 0 so c2 = -80

h = 40 t +80(e^-(t/2) -1)

now when t = 10

h = 400 + 80 (.006-1) = 320

so NOW we have a brand new problem at t = 10 seconds

a = -16

Vo = 39.7 ft/s

Ho = 320 ft

Now

v = vo - at

at the top, v = 0

0 = 39.7 - 16 t

t = 2.48 s

SO part a)

t+the original ten seconds = 12.48 s

h = Ho + Vo t + (1/2) a t^2

h = 320 + 39.7 * 2.48 -8 *2,48^2

h = 320 + 98.5 - 49.2

So part b)

h = 369 feet at the top.

c it falls from 369 feet

v = 0 - 32 t

-369 = (1/2)(-32) t^2

-369 = -16 t^2

t = 4.80 seconds falling

SO

Part c

v at ground = -32(4.8) = - 154 ft/s

I can not do part d because they had g wrong, 16 instead of 32 during the ballistic phase.

If it were -32, as it should be, then

h = Ho + Vo (t-10) + (1/2)(-32) (t-10)^2

Ho is 320

Vo is 39.7 If I did the first ten seconds right

I used t-10 because I assume they mean total t since liftoff, not time after ten seconds.

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