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

Question

Noe Quinn

Physics

2 April, 05:26

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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Answers (1)

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Stanley Banks · 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