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

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Homework Help Overview

The problem involves two balls: one thrown upwards from a building and another dropped from the same height one second later. The objective is to determine the height of the building such that both balls hit the ground simultaneously, given an initial velocity of 6.0 m/s for the first ball.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the equations of motion applied to both balls, questioning the setup and the choice of positive direction for velocity. There are considerations regarding potential rounding errors and variations in the acceleration due to gravity used in calculations.

Discussion Status

Some participants have offered suggestions for checking calculations and considering different values for gravitational acceleration. There is an ongoing exploration of the implications of the chosen sign convention for velocity.

Contextual Notes

Participants note discrepancies in the calculated height of the building compared to an expected answer, prompting discussions about possible errors in calculations or assumptions made in the problem setup.

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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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