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?
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.