A test rocket is launched by accelerating it along a 200.0-m incline at 2.24 m/s
ID: 1528652 • Letter: A
Question
A test rocket is launched by accelerating it along a 200.0-m incline at 2.24 m/s2 starting from rest at point A (the figure (Figure 1) .) The incline rises at 35.0 above the horizontal, and at the instant the rocket leaves it, its engines turn off and it is subject only to gravity (air resistance can be ignored).
Part A
Find the maximum height above the ground that the rocket reaches.
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Part B
Find the greatest horizontal range of the rocket beyond point A.
Figure 1 of 1
A test rocket is launched by accelerating it along a 200.0-m incline at 2.24 m/s2 starting from rest at point A (the figure (Figure 1) .) The incline rises at 35.0 above the horizontal, and at the instant the rocket leaves it, its engines turn off and it is subject only to gravity (air resistance can be ignored).
Part A
Find the maximum height above the ground that the rocket reaches.
hmax = mSubmitMy AnswersGive Up
Part B
Find the greatest horizontal range of the rocket beyond point A.
L = mFigure 1 of 1
200.0 m 35.00Explanation / Answer
Given data:
d=200m
a=2.24m/s^2
angle = 35 degrees
d = 1/2at²
200 = (0.5)(2.24)t^2
t = 13.36 s (time of acceleration along incline)
Velocity(V) of rocket as it leaves incline = (13.36)(2.24) = 29.92m/s
Vertical componet of V = Vy = 29.92 sin 35 = 17.16 m/s
Horizontal componet of V = Vx = 29.92 cos 35 = 24.5 m/s
h = height of incline = 200 sin 35 = 114.7 m
L = horizontal base length of incline = 200*cos 35 = 163.8 m
Time after leaving incline to reach max height = 12.8/g = 17.16/9.8 = 1.75 s
H = 1/2gt² = (0.5)(9.8)(1.75)^2 = 15 m (max vertical height above incline)
Total time after leaving incline to fall to launch height = (2)(1.75) = 3.5 s
Horizontal distance traveled after leaving incline = (24.5)(3.5) = 85.75 m
a) max height above ground = 114.7+ 15 = 129.7 m
b) range of rocket from start = 163.8 + 85.75 = 249.55 m
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