A man drops a baseball from the edge of a roof of a building. At exactly the same time, another man pitches a baseball vertically up toward the man on the roof in such a way that the ball just barely reaches the roof. Does the ball from the roof reach the ground before the ball from the ground reaches the roof, or is it the other way around?
An object is in free fall as soon as it is released whether it is dropped from rest, thrown up, or thrown down so I believe they will touch the roof and the ground at the same time.
Any other suggestions?
2007-01-07 10:13:42 · 4 answers · asked by SMS 1 in Science & Mathematics ➔ Physics
You are correct.
We know this because of the 5 equations of motion.
2nd ball
a= -9.8 m/s^2
d=h
vf=0
t=?
d=vft-1/2at^2
h=(-1/2)(-9.8)t^2
t=sqrt(h/4.9)
1st ball
vi=0
a=9.8
d=h
t=?
d=vit+1/2at^2
h=(1/2)(9.8)t^2
t=sqrt(h/4.9)
As you can see, the times will both be the same.
I hope that this helps.
2007-01-07 10:24:18 · answer #1 · answered by Anonymous · 0⤊ 0⤋
In order for the first ball to hit the ground and the second ball reaches the same high where the first ball is dropped, the final velocity of the first ball has to be the same as the initial velocity of the second ball.
2007-01-07 10:21:11 · answer #2 · answered by 7 · 0⤊ 0⤋
the first ball will reach the ground first
the second ball is pitched so it will first move up until it becames a free falling object, it will then move down again.
Thus it takes more time to reach the ground than the first one
2007-01-07 10:20:03 · answer #3 · answered by James Chan 4 · 0⤊ 2⤋
i agree with you, there will be no winner, clearly they both will hit their targets at the same time.
2007-01-07 10:21:54 · answer #4 · answered by jared kandlstorfer 2 · 1⤊ 0⤋