- #1
Turkish
- 34
- 0
Hi again.
I have 3 sets of questions which I spose are fairly easy for the ones here.
1) A tennis ball of mass 0.20kg is released from rest at a height of 2.0m above a concrete floor.
It rebounds to a height of 1.5m
a) Calculate its kinetic energy and speed just before impact
b) calculate its kinetic energy and speed just after impact
c) calculate its loss of energy between release and its maximum height after rebounding
I have managed to solve a) I did, V^2 = U^2 + 2as
So, 0^2 + 2*10*2.0 = 6.3 So the Velocity is 6.3m/s
Therefore, Kinetic energy = 1/2 mv^2
So, 0.20 * (6.3*6.3 = 40) / 2 = 4J
But I don't understand how I can do b and c, can someone please help me :) Thanks in advance.
I have 3 sets of questions which I spose are fairly easy for the ones here.
1) A tennis ball of mass 0.20kg is released from rest at a height of 2.0m above a concrete floor.
It rebounds to a height of 1.5m
a) Calculate its kinetic energy and speed just before impact
b) calculate its kinetic energy and speed just after impact
c) calculate its loss of energy between release and its maximum height after rebounding
I have managed to solve a) I did, V^2 = U^2 + 2as
So, 0^2 + 2*10*2.0 = 6.3 So the Velocity is 6.3m/s
Therefore, Kinetic energy = 1/2 mv^2
So, 0.20 * (6.3*6.3 = 40) / 2 = 4J
But I don't understand how I can do b and c, can someone please help me :) Thanks in advance.