Help me solve this problem please and thank you. A catapult launches a test rock
ID: 1651566 • Letter: H
Question
Help me solve this problem please and thank you.
A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.2 m/s at ground level. The engines then fire, and the rocket accelerates upward at 4.10 m/s^2 until it reaches an altitude of 960 m. At that point its engines fail, and the rocket goes into free fall, with an acceleration of -9.80 m/s^2. (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? km (c) What is its velocity just before it hits the ground? m/sExplanation / Answer
u= initial velocity = 80.2 m/s
a acceleration= 4.10 m/s2
altitude = 960 m
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a) from S= u*t + (1/2)*a*t2
960 = 80.2*t + (1/2)*4.10*t2
2.05 t2 +80.2 t - 960 = 0
solving this equation , t ( time for rocket to react at an altitude of 960m) = 9.69 s
final velocity (V) at that time, V= u+at
V = 80.2 + 4.10*9.69
V= 119.92 m/s
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time (t') to reach the altitude above 960m
from V= u+a*t'
119.92 = 0 + 9.8*t'
t' = 119.92 / 9.8 = 12.23 s
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altitude above 960 m, h = ?
from V2 - u2 = 2*a*S
119.922 - 0 = 2*9.8*h
h = 733.7 m
max altitude ( total height above ground) = 960+ 733.7 = 1693.7 m ---------- ans->(b)
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time to free fall (t'')=?
from S = ut + (1/2)*a*t''2
1693.7 = 0+ (1/2)*9.8*t''2
t'' = 18.59 s
total time = t+t'+t''
= 9.69 + 12.23 + 18.59 = 40.51 s -------------ans-> (a)
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c) velocity (v) just before hit the ground:
from v2 - u2 = 2*a*s
v2 = 2*9.8*1693.7
v = 182.19 m/s
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