How Do You Calculate the Initial Speed to Throw a Ball Downward from a Building?

[negation]

Homework Statement

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.

Homework Equations

None

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?

 

[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.

 

[negation]

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.

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

 

[TSny]

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]

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.)

As for ball dropped from H:

H = vit + 0.5at^2
H = t(vi + 0.5at) = vi + 0.5at

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?

 

[negation]

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.)

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

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?

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]

 

[TSny]

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

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]

Sounds good!

 

[negation]

Sounds good!

Solved