72. A catapult launches a test rocket vertically upward from a well, giving the

Question

72. A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.0 m/s at ground level. The engines then fire, and the rocket accelerates upward at 4.00 m/s until it reaches an altitude of 1 000 m. At that point, its engines fail and the rocket goes into free fall, with an acceleration of 9.80 m/s2. (a) For what time interval is the rocket in motion above the ground? (b) What is its maximum altitude? (c) What is its velocity just before it hits the ground? (You will need to consider the motion while the engine is operating and the free-fall motion Separately.)

Explanation / Answer

y = v0t + 0.5at²

1000 = 80t + 0.5*4*t²
0 = 2t² + 80t - 1000

from quadratic equations and eliminating the negative answer
t = 10 s to engine cut-off

the velocity at that time is
v = v0 + at
v = 80 + 4(10)
v = 120 m/s

it rises for an additional time
v = gt
t = v/g
t = 120 / 9.8
t = 12.244 s

gaining more altitude
y = 0.5vt
y = 120(12.244) /2
y = 734.64 m

for a peak height of
y = 734.64 + 1000
b) y = 1734.64 m

the time it takes to fall that distance is
y = 0.5gt²
t = sqrt(2y/g)
t = sqrt(2(1734.64)/9.8)
t = 18.815 s

total time in the flight above ground
t = 10 + 12.244 + 18.815
a) t = 41.059 s

v = gt
v = 9.8(18.815)
c) v = 184.387 m/s

Incase of any doubt,please comment

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