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.

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Straight line motion problem:

**Physics Forums | Science Articles, Homework Help, Discussion**