Question

9. A 2-kg ball is dropped out of a w is dropped out of a window from a height of 12 meters (Point A). It hits the grow and bounces back up to a height of 8 meters above the ground (Point). a. What is the potential energy of the ball at points A and C b. What is the velocity of the ball just before and just after it hits the ground (points B and B')? motor to point B It is then

please show work!

Answer 1

a)

At point A,
Potential energy is given by mgh = 2 * 10 * 12 = 240 J ( taking g = 10 m/s^2) or 235.2 J ( taking g = 9.8m/s^2 )

At point C,
Potential energy is given by mgh = 2 * 10 * 8 = 160 J ( taking g = 10 m/s^2) or 156.8 J ( taking g = 9.8m/s^2 )

b)

Just before it hits the ground,
it is the Final velocity(say v ) for motion from point A to B,
Energy is conserved for points A and B,
mgh (potential energy at point A) = 1/2 m v^2 (kinetic energy at point B)
mgh = 1/2 mv²
gh = 1/2 v²
v =
vagh.
=
V2* 9.8 * 12
[ taking g = 9.8m/s^2 ] =  15.33 m/s

After it hits the ground, there is some energy loss during the collision,

Kinetic energy at point B' = Potential energy at point C
1/2 * m * v'² = mg*8
1/2 v'² = 9.8 * 8 [ taking g = 9.8m/s^2 ]
v'² = 2 * 9.8 * 8
v' = 12.52 m/s