a tennis ball bounces on the floor three times. if each time loses 22.0% of its energy due to heating, how high does it rise after the third bounce, provided we released it 2.3 m from the floor?
a.) 140 cm
b.) 110 mm
c.) 110 cm
d.) 11 cm
Total energy = 1 unit
Balance after first collision = 0.78 unit
Balance after second collision = 0.78 ^2 unit
Balance after third collision = 0.78 ^3 = 0.475unit
0.475 (mgh) = mg h”
h” = 0.475*230 cm = 109.25 cm
Answer is 110 cm
The ball has potential energy at any given height.
POTENTIAL ENERGY equation is –> mgh
-m = mass of ball (kg)
-g = gravitational acceleration (m/s^2)
-h = height of ball above surface (m)
Substituting the values into the equation gives,
-m = m (constant)
-g = g (constant)
-h = 2.3m
–>Therefore potential energy is equal to 2.3gm Joules
The ball bounces 3 times and loses 22% or 0.22 of its energy for each bounce, therefore the remainder of its energy will be 78% or 0.78:
1st bounce – 0.78 x potential energy of ball = 0.78 x 2.3gm = 1.79gm Joules
2nd bounce – 0.78 x potential energy of ball after 1st bounce = 0.78 x 1.79gm = 1.40gm Joules
3rd bounce – 0.78 x potential energy of ball after 2nd bounce = 0.78 x 2.3gm = 1.09gm Joules
–>Therefore the energy remaining after the 3rd bounce is 1.09gm Joules
Thus to find its height, let the potential energy equation equal 1.09gm Joules, such that:
– mgh = 1.09gm
-(rearrange the equation to isolate h)- h = 1.09gm/(gm)
–(cancel constants)– h = 1.09m
THEREFORE h ~1.10m = 110cm, THUS THE ANSWER IS ‘C’!!!
schach7 hit the nail on the head. Using energy equations makes this problem easy. (But you missed the .5 in the kinetic energy part). If you did want to do it with laws of motion, you should get the same answer. Using the veloicty equations: v^2 = 2(9.8)(4.93) v = 9.83 v^2 = 2(9.8)(2.68) v = 7.25 meaning a change in velocity of 17.1 meters per second squared, over a time of .00123 seconds, giving 13,900 m/s^2 in the up direction.