Question

In the first stage of a two-stage rocket, the rocket is fired from the launch pad starting from rest but with a constant acceleration of 3.50 m/s2m/s2 upward. At 25.0 ss after launch, the second stage fires for 10.0 ss, which boosts the rocket’s velocity to 132.5 m/sm/s upward at 35.0 ss after launch. This firing uses up all the fuel, however, so after the second stage has finished firing, the only force acting on the rocket is gravity. Air resistance can be neglected.A. Find the maximum height that the stage-two rocket reaches above the launch pad.?
B. How much time after the stage-two firing will it take for the rocket to fall back to the launch pad?
C. How fast will the stage-two rocket be moving just as it reaches the launch pad?

Answer 1

Answer:

a) $h=1991.25\ m$

b) $t=20.1588\ s$

c) $v=197.5563\ m.s^{-1}$

Explanation:

Given:

• acceleration in the first stage of the rocket, $a_1=3.5\ m.s^{-2}$

• duration of the first stage firing, $t_1=25\ s$

• duration of the second stage firing, $t_2'=10\ s$

• velocity at the end of the second stage, $v_2=132.5\ m.s^{-1}$

Now velocity at the end of the first stage:

$v_1=u_1+a_1.t_1$

where

$u_1=$ initial velocity in at the start of the stage 1(at rest) = 0

$v_1=0+3.5\times 25$

$v_1=87.5\ m.s^{-1}$

A)

distance travelled at the end of the first stage:

$s_1=u_1.t_1+\frac{1}{2} \times a_1.t_1^2$

$s_1=0+0.5\times 3.5\times 25^2$

$s_1=1093.75\ m$

acceleration in the second stage:

$v_2=u_2+a_2.t_2$

here:

$u_2=v_1 =87.5\ m.s^{-1}$ (final velocity of the first stage is the initial velocity for the second stage)

$132.5=87.5+a_2\times 10^2$

$a_2=0.45\ m.s^{-2}$

distance travelled in the second stage of the rocket:

$s_2=u_2.t_2+\frac{1}{2} \times a_2.t_2^2$

$s_2=87.5\times 10+0.5\times 0.45\times 10^2$

$s_2=897.5\ m$

Therefore the maximum height reached by the rocket:

$h=s_1+s_2$

$h=1093.75+897.5$

$h=1991.25\ m$

B)

Using eq. of motion:

$h=u.t+\frac{1}{2} \times g\times t^2$

here;

u = initial velocity of the rocket in the event of coming down after reaching the topmost height = 0

$1991.25=0+0.5\times 9.8\times t^2$

$t=20.1588\ s$

C)

The final velocity of the rocket when going down towards the launchpad:

$v^2=u^2+2\times g\times h$

$v^2=0^2+2\times 9.8\times 1991.25$

$v=197.5563\ m.s^{-1}$