Question

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

Answer 1

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 velocity$u=150 m/s$  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

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