Why is my calculation for the height of the building off?

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The discussion centers on a physics problem involving two balls, one thrown upward and the other dropped, to determine the height of a building. The calculations for the height yielded a result of 0.4051 m, while the expected answer is 0.411 m, prompting questions about potential errors. Participants suggest possible causes for the discrepancy, including rounding errors and variations in the gravitational constant used. Additionally, there is clarification on the sign convention for the initial velocity of the thrown ball, emphasizing that if downward is positive, the initial velocity should be negative. The conversation highlights the importance of careful calculations and consistent use of units in physics problems.
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Homework Statement



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?

Homework Equations



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

t2=t1-1seconds


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?
 
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Toranc3 said:

Homework Statement



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?

Homework Equations



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

t2=t1-1seconds


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?

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?
 
PeterO said:
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?

Yeah your right. Thanks buddy!
 
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.
 
HallsofIvy said:
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.

I cat and pasted from original

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

he had -6 ??
 
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