# Ball thrown straight up

1. Jan 28, 2013

### Toranc3

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

A ball is thrown straight up from the edge of the roof of a building. A second ball is dropped from the roof 1 second later. You may ignore air resistance

What must the height of the building be for both balls to reach the ground at the same time if vo is 6.0m/s?

2. Relevant equations

y=yo+vo*t+1/2*a*t^(2)

t2=t1-1seconds

3. The attempt at a solution

I chose positive going down

ball 2:

y=yo+vo*t +1/2a*t^(2)

y=4.905m/s^(2)*t2^(2)
y=4.905m/s^(2)*(t1-1s)^(2)

y=4.905m/s^(2)t1^(2)-9.81m/s*t1+4.905m

ball 1

y=yo+vo*t +1/2a*t^(2)

y=-6m/s*t1 + 4.90m/s^(2) *t1^(2)

Set the final equations equal to each other:
y=y

4.905m/s^(2)t1^(2)-9.81m/s*t1+4.905m=-6m/s*t1 + 4.90m/s^(2) *t1^(2)

4.905m=3.81m/s*t1

t1=1.2874 seconds

plug t1 back into equation for ball 1

I get 0.4051 m but the answer is 0.411 m

What did I do wrong?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jan 28, 2013

### PeterO

I have not checked your calculations closely but a couple of things you may look for.

Your answer is very close, I wonder if there is:
(a) some rounding errors?
(b) the setter used g = 9.8 rather than 9.81?
(c) the setter used g = 10 rather than 9.81?

3. Jan 28, 2013

### Toranc3

Yeah your right. Thanks buddy!

4. Jan 28, 2013

### HallsofIvy

Staff Emeritus
You say in your post "I chose positive going down". If you do that, and the ball were thrown up, $v_0$ must be negative.

5. Jan 28, 2013

### PeterO

I cat and pasted from original

y=-6m/s*t1 + 4.90m/s^(2) *t1^(2)

he had -6 ??

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook