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

a)
`h_m=74625\ m`

b)
`t=265.55\ s`

Step-by-step 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* 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* 30* 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* 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* 9.8* 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+(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* 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`