A 200 kg weather rocket is launched is fired straight up. It starts at rest and

Question

A 200 kg weather rocket is launched is fired straight up. It starts at rest and accelerates upward at 30 m/s2 for 30 s and then runs out of fuel. It subsequently continues upward and drops to the ground after reaching a maximum height. Ignore any air resistance effect. To help you set up the problem, you can assign variables for the height, velocity and time at the launch point as y0, v0, t0, the point where it runs out of fuel as y1, v1, t3.

A. What is the velocity of the rocket when it runs out of the fuel?

B. What the height at which it runs out of fuel?

C. What is the rocket’s maximum height and when does it reach this height?

Explanation / Answer

Here ,

t0 = 30 s

mass of weather , m = 200 Kg

acceleration , a = 30 m/s^2

a) for the final velocity is v1

v1 = a * t

v1 = 30 * 30 m/s

v1 = 900 m/s

the velocity of the rocket when it runs out of the fuel is 900 m/s

b)

for the velocity

Using second equation of motion

y1 = 0.5 * a * t^2

y1 = 0.5* 30 * 30^2

y1 = 13500 m

the height at which the fuel runs out is 13500 m

c)

maximum height = y1 + v1^2/(2*g)

maximum height = 13500 + 900^2/(2 * 9.8)

maximum height = 54826 m

the maximum height of the rocket is 54826 m

time taken when it reach the maximum height = 30 + 900/9.8

time taken when it reach the maximum height = 121.8 s

the time taken when it reach the maximum height is 121.8 s

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