You have designed a test rocket for SpaceX which is launched by accelerating up

Question

You have designed a test rocket for SpaceX which is launched by accelerating up a 250 meter incline that rises 50 degrees above the horizontal. Your rocket accelerates at 1.65 m/s2 starting from rest. As soon as your rocket leaves the end of the ramp, the engines are designed to shut off and the rocket flies through the air only under the influence of gravity. Elon Musk wants to know the following about your design: (A) What is the maximum height your rocket will go? (B) How far away from the start of the ramp will the rocket land? (C) How long does the whole flight take (from lift off to landing)? Can anyone show me how to solve these questions?Please show your wor s)

Explanation / Answer

A) acceleration 1.65 m/s^2 work for 250 meter starting from rest.

a = 1.65 m/s^2 , u = 0 , d = 250m

using, v^2 - u^2 = 2ad

v^2 - 0 = 2(1.65)(250)

v = 28.72 m/s

after ramp ends

initial velocity will be 50 deg above the horizontal.

now vertical component = 28.72sin50= 22 m/s

rocket will go up until its vertical velocity becomes zero.

now a = -9.81 m/s^2 and vf = 0

0^2 - 22^2 = 2(-9.81)(h)

h = 24.68 m

maximum height = 250sin50 + 24.68 = 216.19 m

b) at the end of ramp, rocket is 250sin50 = 191.51 m high.

using y =uy*t + ay*t^2/2

-191.51 = 22t - 9.81t^2/2

4.905t^2 - 22t - 191.51 = 0

t = 8.88 s

horizontal distance traveled in this time = (28.72 cos50) 8.88

    x1 = 163.93 m

horizontal distance due ramp, x2 = 250cos50 = 160.70 m

total horizontal distance = x1 +x2 = 324.63 m

C) time for ramp journey,

using v= u + at

28.72 = 0 + 1.65t

t = 17.41s

total time of flight = 17.41 + 8.88 = 26.29 sec

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