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Kinematics and Dynamics Question: Speed

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A ball is released from top of 98 m tall building. At the same instant a second ball is thrown straight up from the ground with an initial speed of 28 m/s directly below the point where the first ball was dropped. Ignore air resistance.

a. At what height above the ground do the balls collide?

b. What are the balls velocities (magnitudes and directions) just before the collision?

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Solution Summary

The solution determines at what height above the ground do the balls collide.

Solution Preview

Suppose after t seconds and height of h, the collision occurs.

For the dropping ball, 98-h=1/2*g*t^2. (I)
For the rising ball, h=28t-1/2*g*t^2. ...

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