A catapult launches a test rocket vertically upward from a well, giving the rock
ID: 3278915 • Letter: A
Question
A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.4 m/s at ground level. The engines then fire, and the rocket accelerates upward at 4.20 m/s2 until it reaches an altitude of 1030 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?
. s
(b) What is its maximum altitude?
(c) What is its velocity just before it hits the ground? (Assume the positive direction is upward. Indicate the direction with the sign of your answer.)
The response you submitted has the wrong sign. m/s
Explanation / Answer
a]
u = 80.4 m/s
a = 4.2 m/s2
H = 1030 m
velocity at H = 1030m will be:
v = [u2 + 2aH]1/2 = [(80.4)2 + 2(4.2)(1030)]1/2 = 122.947 m/s
and v = u + at
=> 122.947 = 80.4 + 4.2t
=> t = 10.13s
from here on, the acceleration = - 9.8 m/s2
so, at the highest point, v = 0 m/s
=> 0 = 122.947 - 9.8t'
=> t' = 12.545s
therefore, time to reach the highest point will be: T = t + t' = 22.675s
and the additional distance which it covers while decelerating will be:
h = (v2 - u2)/2g = - (0 - 122.9472)/(2 x 9.8) = 771.22m
so, maximum altitude = h' = H + h = 1030 + 771.22 = 1801.22m [answer to (b)]
now, from this height, the rocket makes a free fall
the velocity at the ground will be:
(1/2)mv2 = mgh'
=> v = 187.893 m/s
since the positive direction is upwards, the final velocity will be: v = - 187.893 m/s
this is the velocity with which it hits the ground [answer to (c)]
v = u + at''
187.893 = 0 + 9.8t''
=> t'' = 19.173s
therefore, total time of flight = T + t'' = 22.675 + 19.173 = 41.848s [answer to (a)].
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