I am having a problem solving this:(adsbygoogle = window.adsbygoogle || []).push({});

A ball is thrown straight up from the edge of a 20m tall building (ignore air resistance). One second later a second ball is thrown. What must be the initial velocity of the second ball in order for both of them to hit the ground at the same time.

Knowns:

Ball 1

x=20m

v0=0m/s

a=g (I am using 9.8m/s^2)

t-down=2.02s (x=vt+0.5at^2, 20=0t+0.5(9.8)t^2)

Ball 2

x=20m

v0 is the unknown, t is also unkown

a=g

I don't know how to calculate t-up (ball was thrown up originally) thus I don't know the total time.

The answer in the book is 8.18m/s and t is 1.55s. Any help would be appreciated as to how to obtain those answers.

Using the displacement formula:

x=x0+v0+0.5at^2 I can calculate T-Down for ball #1 which is 2.02s (20=0+0+4.9t^2)

I thought maybe to calculate time down for ball #2 I can subtract 1 second from 2.02s(ball 1 t down) which is 1.02s.

Then substitute that into 20=v0(1.02)+4.9(1.02^2) = 14.6m which is not the correct answer. I assume the error is due to me missing t-up completely.

I also tried solving the equations simoltaneously since they are equal but it seems to make little sense when I do (and I also have 2 unknown on 1 side):

20=v0t+4.9at^2, v0t+4.9at^2-20=0

20=4.9at^2, 4.9at^2-20=0

V0t+4.9at^2-20=4.9at^2-20 (the -20 just cancels and the result is v0t equals 0.

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# Homework Help: Straight line motion problem:

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