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?
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.
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
But why not 9.95/1.33?
it asked for the speed coming BACK
it only came BACK 6.15 meters
it lost kinetic energy in the crash.
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
Why does loss of energy at contact does not effect the vertical time or speed
?Can you explain.
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