Asked by sabeen

anyone who knows the answer to this question please show your work.

drwls i know i posted this queestion before but you didn't reply to my comment.

Your friend has climbed a tree to a height of 6.00 m. You throw a ball vertically up to her and it is traveling at 5.00 m/s when it reaches her. What was the speed of the ball when it left your hand if you released it at a height of 1.10 m?
Answered by bobpursley
Initial KE= change in PE+finalKE
1/2 m vi^2=mgh + 1/2 m vf^2 where h is the difference in heights, ie (6.00-1.10)m
solve for vi
solve for vi
Answered by sabeen
i still don't understand:( what is mgh and what is PE??? what is m? i don't understnad many of the variables in this equation. i was never taught this in class neither have i ever heard of this equation. im sorry mr.pursley. i really am.
Answered by bobpursley
I would drop your class, you are being cheated.

M mass
g acceleration due to gravity
h distance, or change in height
mgh is potential energy (PE)

If these are not in your text, or never had this in class, or have never heard of it, something is wrong. Either you are lost, or the class is going to la la land, at any rate, you are not learning physics.
Answered by sabeen
idk somehow i am doing well.
Answered by Henry
d = 6.0 - 1.1 = 4.9m.

Vf^2 = Vo^2 + 2gd = 5^2,
Vo^2 + 2(-9.8)4.9 = 25,
Vo^2 - 96.04 = 25,
Vo^2 = 25 + 96.04 = 121.04,

Vo = sqrt(121.04) = 11m/s.
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