A small rocket is launched vertically, attaining a maximum speed at burnout of 1.0 ✕ 10^2 m/s and thereafter coasting straight up to a maximum altitude of 1494 m. Assuming the rocket accelerated uniformly while the engine was on, how long did it fire and how high was it at engine cut off

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Physics

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7 February, 07:38

A small rocket is launched vertically, attaining a maximum speed at burnout of 1.0 ✕ 10^2 m/s and thereafter coasting straight up to a maximum altitude of 1494 m. Assuming the rocket accelerated uniformly while the engine was on, how long did it fire and how high was it at engine cut off

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Jamiya · Answer 1

The first thing we can compute for is the altitude the rocket was at engine cutoff using the formula:

v^2 = v0^2 + 2ad

where v = final velocity = 0, v0 = 1 x 10^2 m/s, a = - 9.8 m/s^2, so d is:

- (1 x 10^2 m/s) ^2 = 2 * (-9.8 m/s^2) * d

d = 1,020.41 m

Then finding for the time using the formula:

2d = (v - v0) * t

where d = 1020.41 m, v0 = 0, v = 1 x 10^2 m/s

t = 2 * 1020.41 m / 1 x 10^2 m/s

t = 20.41 seconds