Question #2803

A ball is thrown vertically upward with un initial speed of 20 m/s .Two second later a stone is thrwon 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 ?

Expert's answer

Assume that the release level is at y = 0.

Then the y-coordinate of a ball is changed by the following formula:

y = 20 t – ( 9.8 * t^{2} ) /2

while for the stone its y-coordinate changes as follows:

y = 24(t-1) -( 9.8 * (t-1)^{2} ) /2 = 24t – 24 – (9.8*t^{2})/2 + 9.8 t – 9.8/2 = -28.9 + 33.8 t – (9.8*t^2)/2

First we will find t for which the ball and stone pass each other:

20 t – ( 9.8 * t^2 ) /2 = -28.9 + 33.8 t – (9.8*t^2)/2

20 t = -28.9 +33.8 t

13.8 t = 28.9

t = 28.9 / 13.8 ≈ 2.094

Hence the height above the release point will the ball and stone pass each other is equal to

20 *2.094 – ( 9.8 * 2.094^2 ) /2 = 41.88 – 21.4857 = 20.3943

Answer: h= 20.3943.

Then the y-coordinate of a ball is changed by the following formula:

y = 20 t – ( 9.8 * t

while for the stone its y-coordinate changes as follows:

y = 24(t-1) -( 9.8 * (t-1)

First we will find t for which the ball and stone pass each other:

20 t – ( 9.8 * t^2 ) /2 = -28.9 + 33.8 t – (9.8*t^2)/2

20 t = -28.9 +33.8 t

13.8 t = 28.9

t = 28.9 / 13.8 ≈ 2.094

Hence the height above the release point will the ball and stone pass each other is equal to

20 *2.094 – ( 9.8 * 2.094^2 ) /2 = 41.88 – 21.4857 = 20.3943

Answer: h= 20.3943.

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