- #1
minhngo
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Hi, I appreciate it if someone can reply and help me out with this problem.
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. Air resistance may be ignored.
(a) If the height of the building is H m, what must be the initial speed of the first ball if both are to hit the ground at the same time?
So far I understand that there is:
X(t)=H+Vo(T+1)+1/2(-9.8)(T+1)^2 -> first ball
X(t)=H+1/2(-9.8)(T)^2 -> second ball
Then I know that the two formulas equal each other. From there I found T=(Vo+4.9)/(Vo-9.8)
This is where I get stuck.
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. Air resistance may be ignored.
(a) If the height of the building is H m, what must be the initial speed of the first ball if both are to hit the ground at the same time?
So far I understand that there is:
X(t)=H+Vo(T+1)+1/2(-9.8)(T+1)^2 -> first ball
X(t)=H+1/2(-9.8)(T)^2 -> second ball
Then I know that the two formulas equal each other. From there I found T=(Vo+4.9)/(Vo-9.8)
This is where I get stuck.
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