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A freight train is moving at a constant speed of Vt. A man standing on a flatcar

ID: 2259697 • Letter: A

Question

A freight train is moving at a constant speed of Vt. A man standing on a flatcar throws a ball into the air and catches it as it falls. Relative to the flatcar, the initial velocity of the ball is Vib straight up.

a. What are the magnitude and direction of the initial velocity of the ball as seen by a second man standing on the ground next to the track?

b. How much time is the ball in the air according to the man on the train? According to the man on the ground? Mathematically prove your two answers.

c. What horizontal distance has the ball traveled by the time it is caught according to the man on the train? According to the man on the ground?

d. What is the minimum speed of the ball during its flight according to the man on the train? According to the man on the ground?

e. What is the acceleration of the ball according to the man on the train? According to the man on the ground?


Please show work with a picture carefully explaining each steps. I am having some difficulty with relative veloticy.

Explanation / Answer

a)For the man on the train , the initial velocity is only in the vertical direction that is Vib.

For a man on the ground, the velocity will be Vt in forward direction and Vbi in upward direction.

According to him V = Vt i + Vbi j

Magnitude will be sqrt(Vt^2 + Vbi ^2)

Direction angle from the horizontal x

tan x = Vbi/Vt


b) t = 2V/g where V is the vertical component of velocity.

Since V is same for both the observers , t is same for both of them.

t = 2Vbi/g


c)According to the man on the train , the ball did not travel horizontal distance

as only vertical motion was imparted in reference frame of train.

For man on the ground , d = t * V = (2Vbi/g)Vt = (2VtVbi)/g


d)Minimum speed of ball according to man on train is 0 when it has reached the maximum height.

Minimum speed of ball according to man on ground is Vt when it has reached the maximum height.


e)Acceleration will be g = 9.8 m/s^2 for both the observer as none of the reference frame is accelerating.

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