A catapult launches a rocket at an angle of 53.1° above the horizontal with an i

Question

A catapult launches a rocket at an angle of 53.1° above the horizontal with an initial speed of 84 m/s. The rocket engine immediately starts a burn, and for 3.24 s the rocket moves along its initial line of motion with an acceleration of 27.9 m/s2. Then its engine fails, and the rocket proceeds to move in free-fall.

(a) Find the maximum altitude reached by the rocket.
(b) Find its total time of flight.

(c) Find its horizontal range.

I have asked this question twice now, and it has been wrong twice.

Explanation / Answer

Y-component of velocity, vx = 84sin53.1= 84*0.79 = 66.36 m/s

X-component = vcos53.1 = 84*0.6 = 50.4 m/s

a)

let maximum altitude = d

time = 3.24

using equation s = ut + 0.5at^2 we get

s = Vyt - 0.5*g*(t)^2

s = 66.36*3.24 = 0.5*9.81*(3.24*3.24) = 163.51 m

part b)

total time = time till it goes in a projectile motion + time till it reaches ground

total time = 3.24 + t

now rocket at height 163.51 m starts a free fall

it rocket were not burning and going normally maximum height it would have reached = (vsin(53.1))^2/2g = 224.67 m

so clearly it didn't reach it's maximum height so initial velocity when free fall starts will not be zero

calculation of initial velocity

Vy = uy - gt = 66.36 - 9.8*3.24 = 34.61m/s in +y direction

calculation of t:

we know that s = ut + 0.5gt^2

163.51 = -34.61*t + 0.5*9.8*t^2

4.9t^2 - 34.61t - 163.51 = 0

on solving for t we get

t = 10.3s

total time = 3.24 + t = 3.24 + 10.3 = 13.54s

part c)

Horizontal range = vx * 3.24 = 50.4 * 3.24 = 163.29 m

Hope that's the answer you are looking for :)

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.