could someone help me and show me how to do tis physics problem?

Question

a .2 kg ball was released from a height of 10 meters. the ball hit the ground and rebounded to a height of 9 meters. (a) what was the vertical velocity of a ball just before it hit the ground? (b) what was the vertical velocity of the ball when it left the ground? (c) the collision w/ the ground was inelastic because kinetic energy was lost. how much energy was lost to heat and sound in the collision w/ the ground?

Answer 1

Here,

Mass of the ball, m = 0.2 kg
height, h = 10m
gravitational acceleration, g = 9.8 m/s^2
(a)

At the moment the ball hits the ground, the kinetic energy and the potential energy at height h will be equal. So

1/2 mv^2 = mgh [ v = velocity ]
or, v^2 = 2gh = 2 * 9.8 * 10 = 196
so, v = 14 m/s

Velocity of the ball when it hit the ground = 14 m/s (Ans.)

(b)

at H = 9 m height, the potential energy of the ball will be,
mgH = 0.2 * 9.8 * 9 = 17.64 J

The kinetic energy when the ball rebounded of the ground must be equal to this energy. Say when the ball left the ground it had a velocity o fu in the upwards direction. So,

1/2 mu^2 = mgH = 17.64
or, u^2 = 2 * 17.64/m = 2* 17.64 / 0.2 = 176.4
so, u = 13.28 m/s

The veloctiy of the ball when it left the ground = 13.28 m/s (Ans.)

(c)
Energy loss = Potential energy at 10m - Potential energy at 9m
= mgh - mgH
= mg(h-H)
= 0.2 * 9.8*(10 - 9)
= 1.96 J

Energy loss = 1.96 J (Ans.)

Hope this was helpful.

Answer 2

Ignore air resistence

Let's solve for the energy lost

E(lost)=mgh1-mgh2
=mg(h1-h2)=.2(9.8)(2-1)
=1.96 Joules

let's solve for the velocity at the ground

mgh=1/2mv^2

v=sqrt(2gh)=sqrt(2*9.8*10)=14 m/s

v from ground=sqrt(2*9.8*9)=13.3 m/s