1i dont think so vivek
simple equation
85/100(1/2mv^2+mgh)=mgh
as simple as that.......
A ball looses 15% of its KE after it bounces off a concrete slab . What is the speed with which one must throw it vertically down frm a height of 12.4 m to hav it bounce back to the same height??
-
UP 0 DOWN 0 0 6
6 Answers
71Energy decreases, since mass remains constant so velocity must decrease. So can it reach upto same height.? Please tell me.
1
11@ Vivek, have you written something in the post #2, I can't see anything there except that grey line. [2]
49@vivek: if the ball was dropped from rest, then it would never have returned to its original path...but if u look at the question, the ball is projected downwards with an initial velocity
thus on reaching the ground the net energy is mgh plus the initial kinetic energy of the body...
15% of this energy is being lost
thus after collision the ball has 85% of the sum of mgh plus initial kinetic energy and this is again converted to the potential energy of the ball only...
thus the equation provided by varun is achieved!!! :)
that is it if u call it a theoritical justification!!! :)