18 of 22 Constants In the first stage of a two-stage rocket, the rocket is Part

Question

18 of 22 Constants In the first stage of a two-stage rocket, the rocket is Part A fired from the launch pad starting from rest but with a constant acceleration of 3.50 m/s2 upward A 25.0 s after launch, the second stage fires for 100 s which boosts the rocket's velocity to 132.5 m/s upward at 35.0 s 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 Find the maximum height that the stage-two rocket reaches above the launch pad Submit Request.Answer Part B How much time after the stage-two firing will take for the rocket to fall back to the launch pad? Submit Part C How fast wit the stage-two rocket be moving just as it reaches the launch pad? hysi

Explanation / Answer

PART A

Stage1

final velocity at end of (stage 1) = u+at = 0 + 3.5 * 25 = 87.5 m/s

distance travelled in stage1 =s1= ut+1/2at^2 = 0 + 1/2 *3.5* 25^2 = 1093.75 m .......... [ 1st ]

Stage2

total time in stage2 = 35-25 = 10 s

initial velocity = 87.5 m/s

final veocity =132.5m/s

Distance travelled in stage2 = s2 = 1/2 *10 *( 132.5 + 87.5 ) by s = 1/2 *t*(v+u) formula

s2 = 1100 m ....... [ 2nd ]

After stage2 till highest point(at rest)

a= -9.8m/s^2

final velocity = 0

initial velocity = 132.5 m/s

by v^2 = u^2 + 2 * a * s3

0 = 132.5^2 - 2 * 9.8 * s3

19.6 * s3 = 17556.25

s3 = 895.72 m ........ [ 3rd ]

Therefore, maximum height above launchpad(ground) = 1st + 2nd + 3rd

= 1093.75 + 1100 + 895.72 = 3089.47 m

PART A answer = 3089.47 m

PART B

After stage2 till highest point,

time taken to reach highest point can be found out,

final velocity = 0

initial velocity = 132.5 m/s

acceleration = -9.8 m/s^2

v = u + at

0 = 132.5 - 9.8 * t

t = 13.52 s

this is the time taken to reach highest point after stage2

now from highest point back falling back to launch pad[ motion is downwards ] ,

initial velocity = 0 m/s

distance travelled = 3089.47 m

acceleration = 9.8 m/s^2

s = ut+1/2at^2

3089.47 = 1/2 * 9.8 * t^2

t = 12.55 s

This is the time taken from falling under gravity from highest point to launchpad(ground),

TOTAL time taken after stage 2 to launchpad(ground) is

time taken to reach highest point after stage2 + time taken from falling under gravity from highest point to launchpad(ground)

= 13.52 + 12.55 = 26.07 s

Answer to part B is 26.07 s

PART C

From highest point to launchpad

we need to find final velocity at the ground

initial velocity at highest point = 0

final velocity = v (let)

distance travelled = 3089.47 m

acceleration = 9.8m/s^2

v^2 = 0 + 2 * 9.8 * 3089.47

v = 246.07 m/s

Answer to PART C is 246.07 m/s

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