Asked by Raj

h ttps://www.google.com/search?q=velocity+of+ball+after+rebounding+is+4.6&source=lnms&tbm=isch&sa=X&ved=2ahUKEwiAu-nqpfLoAhXMwzgGHeUSD2MQ_AUoAXoECAoQAw&biw=1366&bih=656#imgrc=4JDD8NnQiwxwvM

No space between h and t.
For the velocity immediate after rebounding why can't we take 9.95 as it is distance covered horizontally?

Answers

Answered by oobleck
first, because velocity is m/s and distance is m.

The horizontal velocity is 15 cos60°, which does not change. When it rebounds, it leaves with the same speed that it hit with, right? Unless some of the KE is absorbed by the bounce.
Answered by Damon
It hits at height h going horizontal at peak
initial horizontal speed = 15 cos 60 = 7.5 m/s
time to wall = 9.95/7.5 = 1.33 s
then also takes 1.33 s to fall, same old g
6.15 /1.33 = 4.62 m/s
Answered by Raj
But why not 9.95/1.33?
Answered by Damon
it asked for the speed coming BACK
it only came BACK 6.15 meters
it lost kinetic energy in the crash.
Answered by Damon
The loss of energy at contact does not effect the vertical time or speed
h = (1/2) g t^2
v = g t
however it does cut the horizontal speed
Answered by Raj
Why does loss of energy at contact does not effect the vertical time or speed
?Can you explain.
Answered by Damon
There is no vertical force at contact
it does the usual thing up and down
+Vi (up) at start
a = -g
v =Vi-g t
v = 0 at top
v = 0 - g t
-Vi (down) at bottom
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