Question

A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.6 m/s at ground level. The engines then fire, and the rocket accelerates upward at 3.90 m/s2 until it reaches an altitude of 1040 m. At that point its engines fail, and the rocket goes into free fall, with an acceleration of −9.80 m/s2. (you will need to consider the motion while the engine is operating and the free-fall motion separately.) (a) for what time interval is the rocket in motion above the ground?

Answer 1

solution:

y = v0t + ½at²

1150 = 79t + ½3.9t²

0 = 3.9t² + 158t - 2300

from quadratic equations and eliminating the negative answer

t = (-158 + v158² -4(3.9)(-2300)) / 2(3.9)

t = 11.37 s to engine cut-off

the velocity at that time is

v = v0 + at

v = 79 + 3.9(11.37)

v = 123.3 m/s

it rises for an additional time

v = gt

t = v/g

t = 123.3 / 9.8

t = 12.59 s

gaining more altitude

y = ½vt

y = 123.3(12.59) /2

y = 776 m

for a peak height of

y = 776 + 1150

Answer 2

Answer:

t = 31.52 s, the time until the crash

t = 42 s, time to the motion

Explanation:

We have the following data:

so = 0

vo = 80.6 m/s

a = 3.90 m/s^2

s = 1040 m

replacing in the distance equation we have:

s = so + vo*t + (a*t^2)/2

s = 0 + 80.6*t + (3.9 * t^2)/2 = 1040

clearing t:

t = 10.48 s

v1 = v2o = vo + a*t = 80.6 + (3.9 * 10.48) = 121.472 m/s

s2 = 1040 + 121.472t - (9.8*t^2)/2

0 = 1040 + 121.472t - (9.8*t^2)/2

Clearing t:

t = 31.52 s, the time until the crash

t = 42 s, time to the motion