physics help needed please?

A

Alex J

New member
A ball is thrown horizontally from the top of a building 100 meters high. The ball strikes the ground at a point 65 meters horizontally away from the point of release.

What is the initial horizontally velocity of the balls?

What is the speed of the ball just before it hits the ground?

Please explain how you go the answer :)

Thanks
 
In vertically downward direction:-
Initial velocity = 0 (because the ball is moving horizontally)
Acceleration = g = 9.8 m/s^2
Displacement h = 100 m
h = 0 * t + 1/2 * gt^2
h = 1/2 * gt^2
t = sqrt(2h/g) = sqrt(2 * 100/9.8) = 4.52 s

In horizontal direction:-
Displacement d = 65 m
Velocity = d/t = 65/4.52 = 14.4 m/s

When the ball hits the ground:
The vertical component of the velocity = sqrt(2gh)
= sqrt(2 * 9.8 * 100) = 44.27 m/s
The horizontal component of the velocity = 14.4 m/s
Therefore the speed = sqrt(44.27^2 + 14.4^2)
= 46.5 m/s

Ans: 14.4 m/s, 46.5 m/s
____________________________

Note: Horizontal component of the velocity remains unchanged because there is no net force in the horizontal direction.
 
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