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

Question

A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 79.8 m/s at ground level. The engines then fire, and the rocket accelerates upward at 3.80 m/s2 until it reaches an altitude of 1010 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?

(b) What is its maximum altitude?

(c) What is its velocity just before it hits the ground?

Explanation / Answer

Solution-

y = y0 + v0t + ½at²

1010 = 0 + 79.8 t + ½ 3.80t²
0 = 1.9 t² + 79.8t - 1010

t = 10.19 s

The velocity at engine cutout is

v = v0 + at
v = 79.8 + 3.80(10.19)
v = 118.52 m/s

The time needed to slow from cut out speed to zero is

v = gt
t = v/g
t = 118.52/9.80
t = 12.09 s

in that time it reaches a height of

y = y0 + v0t + ½gt²
y = 1010 + 118.52(12.09) + ½(-9.8)(12.09)²
y = 9461.89 m (ANS (b))

The time needed to fall back to ground from that height is
y = ½gt²
t = (2y/g)
t = (2(9461.89)/9.8)
t = 43.94 s

so total time above the ground is
43.94 + 12.09 + 10.19 = 66.22 s (ANS(a))

The velocity reached just before ground impact

v = gt
v = (-9.80)(43.94)
v = --430.612 m/s (ANS (c))

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