# Homework Help: Dropped ball Kinematics motion

1. Dec 12, 2013

### negation

1. The problem statement, all variables and given/known data

You're atop a building of height, h, and a friend is poised to drop a ball from a window at h/2. Find an expression for the speed at which you should simultaneously throw a ball downward, so the two hit the ground at the same time.

2. Relevant equations

None

3. The attempt at a solution

1)H = vit + 0.5gt^2
t1 = SQRT[(H - vit)/-4.9ms^-2]

2)H/2 = vit + 0.5gt^2
t2 = SQRT[(H/2-vit)/-4.9ms6-2]
t2 is the time taken for ball to fall from H/2 to ground

t2=t1
so,
set t1= SQRT[(H - vit)/-4.9ms^-2] to be equals to t2
therefore,
SQRT[(H/2-vit)/-4.9ms6-2] = SQRT[(H - vit)/-4.9ms^-2]

vi = [H - 4.9ms^-2(h/2-vit)/(-4.9ms^-2)]/t

Could someone give me a let up?

Last edited: Dec 12, 2013
2. Dec 12, 2013

### TSny

Hello.

Note that one of the balls is dropped. For this ball, try to find an expression for the time it takes the ball to reach the ground in terms of H and g.

3. Dec 12, 2013

### negation

If the ball is dropped from H/2, then the assumption vi=0 can be made.
Hence, H/2 = -4.9ms^-2t^2
t = SQRT[(H/2)/-4.9ms^-2]

As for ball dropped from H:

H = vit + 0.5at^2
H = t(vi + 0.5at) = vi + 0.5at
t = [H -vi]/05a = [H - vi]/-4.9ms^-2

4. Dec 13, 2013

### TSny

OK, but it looks like you've run into a sign problem. You have an expression where you're taking the square root of a negative number.

I suggest that you express your results in terms of the symbol g and not substitute a numerical value for g. (Since a value for H is not given, you won't be able to find a numerical answer to the question.)

You can't just drop out the factor of t

Is there any relationship between the t in this equation and the t you got for the ball that was dropped from H/2?

5. Dec 13, 2013

### negation

H/2 = 0.5gt^2
t = SQRT[H/g]

What a terrible elementary blunder.

It appears logical to substitute t = SQRT[H/g] into H = vit + 0.5gt^2. Then isolate vi from the other variables.
The ball thrown from H and the ball dropped from H/2 must both touch the ground at t = SQRT[H/g]
From this, it can be deduced that the distance H = vit + 0.5gt^2 must be achieved in t = SQRT[H/g]

Last edited: Dec 13, 2013
6. Dec 13, 2013

### TSny

Sounds good!

7. Dec 13, 2013

### negation

Solved

Last edited: Dec 13, 2013