Question

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?

Answer 1

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.

Answer 2

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

Answer 3

But why not 9.95/1.33?

Answer 4

it asked for the speed coming BACK
it only came BACK 6.15 meters
it lost kinetic energy in the crash.

Answer 5

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

Answer 6

Why does loss of energy at contact does not effect the vertical time or speed
?Can you explain.

Answer 7

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