Question 428.
A rocket blasts off vertically from rest on the launch pad with an upward acceleration of 2.70 m/s^2 . At 10.0s after blastoff, the engines suddenly fail, which means that the force they produce instantly stops.
How long after it was launched will the rocket fall back to the launch pad?
Answer 428.
The upward velocity of the rocket after 10 s = 2.70 * 10 = 27 m/s
Distance traveled in 10 s = (1/2) * 2.7 * 10^2 = 135 m
Let t = time for the rocket to reach the ground from this height of 135 m
Using s = ut + (1/2) gt^2
with s = - 135, u = 27, g = - 9.81
=> - 135 = 27t - 4.95 t^2
=> 4.95 t^2 - 27t - 135 = 0
=> t = (1/9.81) [27 + √((27)^2 + 2673)]
=> t = 8.7 s
=> time to fall back after launching
= 10 + 8.7 s
= 18.7 s.
Link to YA!
A rocket blasts off vertically from rest on the launch pad with an upward acceleration of 2.70 m/s^2 . At 10.0s after blastoff, the engines suddenly fail, which means that the force they produce instantly stops.
How long after it was launched will the rocket fall back to the launch pad?
Answer 428.
The upward velocity of the rocket after 10 s = 2.70 * 10 = 27 m/s
Distance traveled in 10 s = (1/2) * 2.7 * 10^2 = 135 m
Let t = time for the rocket to reach the ground from this height of 135 m
Using s = ut + (1/2) gt^2
with s = - 135, u = 27, g = - 9.81
=> - 135 = 27t - 4.95 t^2
=> 4.95 t^2 - 27t - 135 = 0
=> t = (1/9.81) [27 + √((27)^2 + 2673)]
=> t = 8.7 s
=> time to fall back after launching
= 10 + 8.7 s
= 18.7 s.
Link to YA!
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