Question

A 200 kg weather rocket is loaded with 100 kg of fuel and fired straight up. It accelerates upward at 30 m/s2 for 35 s , then runs out of fuel. Ignore any air resistance effects.
a. What is the rocket's maximum altitude?
b. How long is the rocket in the air?
c. Draw a velocity-versus-time graph for the rocket from liftoff until it hits the ground.

Answer 1

Answer:

a) $h_m=74625\ m$

b) $t=265.55\ s$

Explanation:

Given:

• mass of rocket, $m_r=200\ kg$

• mass of fuel, $m_f=100\ kg$

• acceleration of the rocket consuming fuel, $a=30\ m.s^{-2}$

• time after which the fuel exhaust, $t_f=35\ s$

During the phase of fuel exhaustion:

velocity attained by the rocket just as the fuel ends:

$v_f=u+a.t_f$

where:

$u=$ initial velocity of the rocket = 0

$v_f=0+30\times 35$

$v_f=1050\ m.s^{-1}$ this will be the initial velocity for the phase of ascending of the rocket's height under the influence of gravity.

height at which the fuel finishes:

$v_f^2=u^2+2a.h_f$

$1050^2=0^2+2\times 30\times h_f$

$h_f=18375\ m$

During the phase of ascend in height of rocket after the fuel is over:

Time taken to reach the top height after the fuel is over:

$v=v_f+g.t'$

at top v = (final velocity during this course of motion )= 0 $m.s^{-1}$

$0=1050-9.8\times t'$

$t'=107.1429\ s$

Height ascended by the rocket after the fuel is over:

$v^2=v_f^2+2g.h'$

at the top height the velocity is zero

$0^2=1050^2-2\times 9.8\times h'$ (-ve sign denotes that the direction of motion is opposite to that of acceleration)

$h'=56250\ m$

Therefore the maximum altitude attained by the rocket:

$h_m=h_f+h'$

$h_m=18375+56250$

$h_m=74625\ m$

b)

time taken by the rocket to fall back to the earth:

$h_m=v.t_m+\frac{1}{2} g.t_m^2$

where:

$v=$ initial velocity of the rocket during the course of free fall from the top height.

$74625=0+4.9\times t_m^2$

$t_m=123.41\ s$

Now the total time for which the rocket is in the air:

$t=t_f+t'+t_m$

$t=35+107.1429+123.41$

$t=265.55\ s$