- #1
Toranc3
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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?