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qwelly [4]
3 years ago
6

A rocket moves upward, starting from rest with an acceleration of +30.0 m/s2 for 5.00 s. It runs out of fuel at the end of this

5.00 s and continues to move upward. How high does it rise? m
Physics
1 answer:
olchik [2.2K]3 years ago
7 0

Answer:

The distance covered by the rocket after fuel ran out is 3442.04 m

Explanation:

Given that the rocket moves with an acceleration a=30m/s^2

time t=5 s

Since the rocket starts from rest initial velocity  u=0 s

The distance it travelled within this time is given by  s=ut+ \frac{1}{2} at^2                                                                                                  =0 \times 5+ \frac{1}{2} (30\times25)=375 m

Velocity at this point is given by v=u+at

v=0+30\times5=150m/s

Given that at this height it runs out of fuel but travels further. Here final velocity v=0(maximum height), initial velocityu=150 m/s  and time to zero velocity t=\frac{v}{g} = \frac{150}{9.8} =15.3 s.

Thus it travels 15.3 seconds more after fuel running out. The distance covered during this period is given

s= ut+\frac{1}{2} gt^2=150 \times 15.3+1/2 \times9.8 \times 15.3^2=3442.04 m

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