Problem one

1. Sep 9, 2008

kmikias

Hi..I have a question I guess you guys help me out ...here is the question.

1.A ball is thrown verticaly upward with an initial speed of 16m/s.then 1.1s later a stone is thrown straight up(from the same inital height of the ball ) with inital speed of 21.3m/s.
the acceleration of gravity is 9.8m/s^2.

how far above the realease point will the ball and the stone pass each other?answer in meter.

Here is what i did.
Y(ball) = Y(stone)

V(inital)t + 1/2 9.8t^2 = V(inital)(t -1.1)+ 1/2 9.8t^2

so 1/2 9.8t^2 will cancel out

which left
16m/s t = 21.3t - 23.43

t = 4.4207 second

after that i use the same formula to find distance which is distance of the ball is 166.4898 and distance of stone is 189.9195

finaly i subtacted stone - ball = 23.4297 meter.

but still my answer is incorrect but i don't know where is my mistake ...............so i just need little help

thank you

2. Sep 9, 2008

Kurdt

Staff Emeritus
Careful with the highlighted time.

3. Sep 9, 2008

kmikias

thank you Kurdt that where i did a mistake.

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