Help with a kinematics problem, please? ?

Question

=] help would be greatly appreciated.

A ball is dropped from a height of 2 meters and rebounds from the floor to a height of 1.5 meters.

- What is the velocity of the ball just before it hits the floor?

- What is its velocity just as it leaves the floor?

- If the ball is in contact with the floor for 0.02 seconds, what are the magnitude and the direction of the average acceleration of the ball while in contact with the floor?

Answer 1

Let u = velocity just before hitting the floor,

v = velocity just after hitting the floor.

Acceleration due to gravity = 9.8 m/s^2

Ball falls from 2 meter

Therefore, u^2 = 2*9.8*2 = 39.2

Or, u = sqrt(39.2) = 6.26 m/s downward

v is upward. g pulls the ball downward. Displacement = 1.5 meter. Velocity after reaching the top = 0

Therefore, v^2 = 2*9.8*1.5 = 29.4

Or, v = sqrt(29.4) = 5.42 m/s upward

Time of contact t = 0.02 second

Average acceleration = (v-u)/t

= (5.42 - (-6.26))/0.02  (note: using -6.26 because u is opposite to v)

= (5.42 + 6.26)/0.02

= 584 m/s^2

This is upward because this acceleration causes the ball to move up.

Ans: 6.26 m/s downward

5.42 m/s upward

584 m/s^2 upward