Physics question involving a rocket...?

Question

Well, I've tried to answer this question five times, but keep getting it wrong. Please help me out!

"A rocket is fired vertically upward from a well. A catapult gives it an initial speed of 80.0 m/s at ground level. Its engines then fire and it accelerates upward at 4.00 m/s^2 until it reaches an altitude of 960m. At this point its engines fail, and the rocket goes into free fall, with an acceleration of -9.80 m/s^2.

A) How long is the rocket in motion above the ground?

B) What is its maximum altitude?

C) What is its velocity just before it collides with the earth?

Any pointers would be greatly appreciated.

Answer 1

(A) Let us start at ground level. The rocket here has an upward speed of 80 m/s and it acclerates at 4 m/s^2 until 960 m.

s = s0 + v0t + (1/2)at^2

s = (1/2)at^2  .... when s0 and v0 = 0

v = v0 + at

The total flight time can be divided into sections:

1. time of accelerated upward flight to 960 m altitude

2. time to coast upward until gravity causes it's speed to decrease to 0

3. time to fall to the earth from this max altitude

1. time of accelerated flight:

s = v0t + (1/2)at^2

960 = 80t + (1/2)(4)t^2

2t^2 + 80t - 960 = 0

t^2 + 40t - 480 = 0

t = [-40 +/- SQRT(1600 + 4(480)]/2 ..... 8 SQRT(55)

t = -20 +/- 4SQRT(55)

t = 9.665 seconds (ignore the negative solution)

2. coasting time. this will end when the rocket has velocity = 0.

v (at burnout) = v0 + at = 80 + 4(9.665)

vbo = 118.66 m/s

v = vbo - (9.8)t .... effect of gravity on velocity after burnout

0 = 118.66 - (9.8)t

t = 118.66/9.8 = 12.108 seconds

3. Timt to fall to earth,

max height = H = s0 + v0t - (1/2)gt^2 = 960 + vbo*t - (1/2)(9.8)t^2

where t is the time determined in (2)

and v0 in this case is the velocity at burnout = vbo from (2)

H = 960 + 118.66(12.108) - (4.9)(12.108)^2

H = 1678.377 meters

H = (1/2)gt^2

H = (1/2)(9.8)t^2

t = SQRT(2*1678.377/9.8)

t = 18.507 seconds

The time the rocket is in motion above the ground is the sum:

total time = 9.665 + 12.108 + 18.507

total time = 37.281 seconds

(B) Maximum altitude = 1678.377 meters .... calculated in (A)

(C) v = -gt = -9.8(18.507) .... time from (3) above

v = -181.4 m/s .... the rocket is heading down