How high does the ball go after bouncing?

Question

A rubber ball with a mass of .210 kg is thrown exactly horizontally form a height of 2meters with an initial spedd of v=13m/s. when it strikes the ground, it bounces off at the same angle it made when it landed. During the bounce, it loses 20% of it's kinetic energy to dissipative forces. How high does it go after bouncing?


I know the answer is 1.6 but I'm still confused as to why....my teacher printed out answers and showed how he did the problem but his way was rather confusing and I would just like to see it done from another perspective if possible...thanks

Answer 1

One way to look at it is :

Ignore the horizontal motion of the ball - it doesnt change
Ignore the mass of the ball - it doesnt change
Ignore all kinetic energy considerations vertically - at the start and top of the bounce it is zero.

When the ball reaches its height after bouncing; it has 80% of its initial potential energy, so nothing else having changed, it must be 80% of its original height.

Answer 2

Actually this question is designed to confuse you. You don't even need the horizontal velocity.
Since the vertical motion of the ball can be treated seperately of the horizonta motion,
Potential Energy (PE) at top is = Kinteic energy at bottom
=mgh
=0.21 x 9.8 x 2
=4.116 J
Now at the bottom 20% is lost so only 80% is left, thus
80% of 4.116
=80/100 x 4.116
=3.2928J
Now again this KE will be equal to the PE at the top of the second bounce
3.2928 = mgh
3.2928 = 0.21 x 9.8 x h
h=1.6m

Answer 3

Usaully about half the distance than dropped from, decreasing the same, with each bounce... i.e. tennis ball