A science student is riding on a flatcar of a train traveling along a straight horizontal track at a constant speed of v1. The student throws a ball along a path that she judges to make an initial angle of θ with the horizontal and to be in line with the track. The student’s professor, who is standing on the ground nearby, observes the ball to rise vertically. How high does the ball rise?.

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AnnuG AnnuG · Answer 1 · 2022-09-20T07:45:38+02:00

The ball rises to a height of s = ( [tex]V_{1}[/tex] tan θ )² / 2g m

[tex]V_{y}[/tex] / [tex]V_{x}[/tex] = tan θ

Here [tex]V_{y}[/tex] is the vertical velocity and [tex]V_{x}[/tex] is the horizontal velocity which is the velocity of train.

[tex]V_{y}[/tex] = [tex]V_{1}[/tex] tan θ ( θ is the angle of projection )

We know that,

a = [tex]V_{y}[/tex] / t

t = [tex]V_{y}[/tex] / a

where,

a = Acceleration due to gravity = g

t = Time

s = ut + 1 / 2 at²

s = 1 / 2 ( g ) ( [tex]V_{1}[/tex] tan θ / g )²

s = ( [tex]V_{1}[/tex] tan θ )² / 2g m

The height of a projectile is the distance of thrown object from the point where it is thrown vertically.

Therefore, the ball rises to a height of s = ( [tex]V_{1}[/tex] tan θ )² / 2g m

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