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Problem

A firework rocket is fired vertically upwards with a constant acceleration of 4 ms^-2 until the chemical fuel expires. Its ascent is then slowed by gravity until it reaches a maximum height of 138 metres.

Assuming no air resistance and taking g=9.8 ms^-2, how long does it take to reach its maximum height?

Solution

During acceleration phase (as fuel burns), a=4, u=0, let v=w.

v=u+at: w=4t₁ (1)
v²=u²+2as: w²=8s₁ (2)

During deceleration phase (fuel expired), a=-9.8, u=w, v=0.

v=u+at: 0=w9.8t₂, w=9.8t₂ (3)
v²=u²+2as: 0=w²19.6s₂, w²=19.6s₂ (4)

Equating (1) and (3), 4t₁=9.8t₂, t₂=(20/49)t₁, so total time to reach maximum height, t₁+t₂=(69/49)t₁.

Equating (2) and (4), 8s₁=19.6s₂, s₂=(20/49)s₁, and as
s₁+s₂=(69/49)s₁=138, we get s₁=98.

Using s=ut+½at² during acceleration phase, s₁=2t₁², t₁=(s₁/2)=7.

Hence time to reach maximum height is (69/49)7=69/7 seconds.