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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 :

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

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