A 5300-kg cart carrying a vertical rocket launcher moves to the right at a const

Question

A 5300-kg cart carrying a vertical rocket launcher moves to the right at a constant speed of 39.0 m/s along a horizontal track. It launches a 42.4-kg rocket vertically upward with an initial speed of 40.1 m/s relative to the cart.

a)How high will the rocket go?

b)Where, relative to the cart, will the rocket land?(in cart/behind cart )<-- choose one of them

c)How far does the cart move while the rocket is in the air?

d)At what angle, relative to the horizontal, is the rocket traveling just as it leaves the cart, as measured by an observer at rest on the ground?

e)What is the rocket's trajectory as seen by an observer stationary on the cart ?(the rocket travels in a cycloid/the rocket travels straight up and then straight down/the rocket travels in a parabola)<-- choose one

"clear answer plz "

Explanation / Answer

a)
Consider vertical motion of rocket:
Vi = 39 m/s
use:
Vf^2 = Vi^2 + 2*a*h
0 = 40.1^2 + 2*(-9.8)*h
h=82 m
Answer: 82 m

b)
When rocket was launched it horizontal velocity with respect to ground was 39m/s
That is same as speed of cart.
So rocket will land into thet cart

c)
Time taken by rocket to reach max height be t'
then Consider vertical motion of rocket:
Vf= Vi+a*t
0 = 40.1 - 9.8*t
t=4.09 s

So total time of flight= 4.09*2 = 8.18 s

Speed of cart= 39 m/s

distance moved by cart = speed * time
= 39m/s * 8.18 s
=319.2 m
Answer:319.2m

d)
Horizontal velocity of rocket= 39m/s
Vertical velocity of rocket= 40.1 m/s

angle = atan (40.1/39) =45.8 degree

e)
As seen from cart,it will goup and fall down.There is no horizontal velocity.
the rocket travels straight up and then straight down

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