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## Homework Statement

A ball is thrown vertically upward with an initial speed of 20 m/s. Two seconds later, a stone is thrown vertically (from the same initial height as the ball) with an initial speed of 24 m/s. At what height above the release point will the ball and stone pass each other?

## Homework Equations

y

_{f}= y

_{i}+ v

_{i}t + ½ at

^{2}

## The Attempt at a Solution

The right solution :

The ball is thrown 2 sec before the rock

yball = ½ at

^{2}+ v

_{i}t + y

_{i}

yball = ½(-10)(t+2)

^{2}+ 20(t+2)

yball = -5t

^{2}+ 20

yrock = ½ at

^{2}+ v

_{i}t + y

_{i}

yrock = ½(-10)t

^{2}+ 24t

yrock = -5t

^{2}+ 24t

yball = yrock (they pass each other)

-5t

^{2}+ 20 = -5t

^{2}+ 24t

t = 5/6 seconds

yrock = -5(5/6)2 + 24(5/6)

yrock = 17 meters

but I don't know why he considered (t) of the ball (t+2) althought the ball is thrown 2 seconds before the stone???

isn't it true if :

t of the ball = t-2

t of the stone = t

?

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