(a) What is the minimum speed of the coinrelative to the ground during its fligh

Question

(a) What is the minimum speed of the coinrelative to the ground during its flight?
1 m/s

At what point in the coin's flight does this minimum speedoccur?
2 immediately after beingtossed
at the top of itsmotion    
immediately before beingcaught

Explain.
3
(b) Find the initial speed and direction of the coin as seen by anobserver on the ground.
4 m/s
5°(counterclockwise from the +x-axis)

(c) Use the expression for ymax = (v0sin)2 2g to calculate the maximum height of the coin (above the trainfloor), as seen by an observer on the ground.
6 m

(d) Calculate the maximum height of the coin (above the trainfloor) from the point of view of the passenger, who sees only theone dimensional motion.
7 m 2 immediately after beingtossed
at the top of itsmotion    
immediately before beingcaught
2 immediately after beingtossed
at the top of itsmotion    
immediately before beingcaught
2 ymax = (v0sin)2 2g (v0sin)2 2g

Explanation / Answer

The minumum speed of thecoin relative to the ground equals that
of the train = 13.1 m / sec
This will occur at the top of its motion because its vertical
speed is zero at that point
On the train vx =0 vy = 3.88 m/s On the ground vx = 13.1 m/s vy =3.88 m/s Relative to the ground v = (3.882 +13.12) = 13.66 m/s Relative to ground tan = 3.88 / 13.1 =.296   and = 16.5 deg Maximum height of coin is the same on the groundand on the train t = vy / g = 3.88 / 9.8 = .396sec    time for coin to reach zero vertical speed(top of trajectory) h = vy t - 1/2 g t2 = 3.88 * .396 - 1/2* 9.8 * .3962 = .768 m on both train and ground I don't see the point in multiplying 13.66 * sin 16.5 = 3.88(we already know that)

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