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A small rocket is launched vertically, attaining a maximum speed at burnout of 1.0 102 m/s and thereafter coasting straight up to a maximum altitude of 1528 m. Assuming the rocket accelerated uniformly while the engine was on, how long did it fire?

and how high was it at engine cutoff?

Verified Answer :

vf^2 = vo^2 + 2ad

vf = 0 m/s
a = -9.8

0  =  10404  + 2 (9.8)d

10404 = 2 (9.8) d

d = 530.816

1528 -  530.816 = 997.184

Hope this helps

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