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 3.90 m/s2 until it reaches an altitude of 1130 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.). What is its velocity just before it hits the ground?

Answer 1

y = v0t + ½at²

1130 = 80.4 t + ½ 3.9 t²
0 = 3.9 t² + 160.8 t - 2260

from quadratic equations and eliminating the negative answer

t = (-160.8 + √160.8² -4(3.9)(-2260)) / 2(3.9)
t = 11.07 s to engine cut-off

the velocity at that time is
v = v0 + at
v = 80.4 + 3.9(11.07)
v = 123.5 m/s

it rises for an additional time
v = gt
t = v/g
t = 123.5 / 9.8
t = 12.60 s

gaining more altitude
y = ½vt
y = 123.5(12.60) /2
y = 778.05 m

for a peak height of
y = 778.05 + 1130
b) ►y = 1908.05 m

the time it takes to fall that distance is
y = ½gt²
t = √(2y/g)
t = √(2(1908.05)/9.8)
t = 19.7 s

total time in the flight above ground
t = 19.7 + 12.60 + 11.07
a) ►t = 43.37 s

v = gt
v = 9.8(19.7)
c) ►v = 193.06 m/s