i need explantion?

Question

a basket ball rolls of the rim and falls to the floor from a hiehg tof 3.05m. then it bounces up and loses 15% of its kinetic energy. To what height will it rise this time?...

my teacher explained it with 4 different equations but i dont get it...

Answer 1

Ok, the ball will have a collision with the floor, right ?, but first you need to calculate the velocity of the ball when it reaches the floor.

The ball falls from 3.05 m

Potential energy = m*g*3.05

g = gravity = 9.8 m/s^2

Kinetic energy when the ball reaches the floor : m*v^2 / 2

By conservation of energy : m*v^2 / 2 = m*g*3.05

v = 7.73 m/s

If the ball loses 15% of its kinetic energy, then, the ball will go up again, with (100 -15)% of it's initial kinetic energy.

But the initial kinetic energy is : m*v^2 / 2

If the ball starts know with :

85/100(m*v^2/2) = m*g*h'

h' = the new height

but we know that : mv^2 / 2 = mg*3.05, then

85/100*m*g*3.05 = mgh'

h' = 2.6 meters

Actually, we didn't need to find the velocity, but I did it becaus you may understand the meaning of the losing energy.

When the ball collides with the floor, it loses energy becaus its speed decreases ok.

Hope that helped

Answer 2

the concept here is that the basketball has a potential energy due to its height that is converted to kinetic energy when it first hits the floor. Then 15% is lost in the collision with the floor then the baskeball bounce up until it's kinetic energy is converted back into potential energy.

PE = mgh
KE = 1/2 mv^2

at rim
PE = m x g x (3.05 meters), KE = 0
at floor
PE = 0, KE = 1/2 mv^2 = PE(at rim) = m x g x 3.05
after collision
PE = 0, and KEnew = .85 x KE = .85 x m x g x 3.05 = m x g x 2.59
at new height
KE = 0, PE = KE after collsion = m x g x 2.59 m

since neither m nor g changed, new height = 2.59 m

Answer 3

2.5925m in the air on the second try.