a bouncing ball of mass 200g leaves the ground with a kinetic energy of 10J
a)if the ball rises vertically, calculate the maximum height it is likely to reach.
b)in practice, the ball rarely reaches the maximum height. explain why this is so.
6 answers
please help please i find this difficult if someone could just explain please
dr bob can you help
help if you can
I'm sorry, but I'm a real dummy in physical science.
do you no anyone who know how to do this, this is due in school
Use kinetic energy
Make sure that if you have Joules, you have kg, not grams.
K = 1/2 mv^2
10 = 1/2 (.2) v^2
10 = .1v^2
100=v^2
10 = v
So, the motion of the ball is
h = 10t - 4.9t^2
max height at t = -10/-9.8 = 1.02
h(1.02) = 5.10
The ball should come pretty close to this height, since we know its initial velocity. The only thing that could affect its height is air resistance. Any inelasticity of the rubber does not come into play, since we are only analyzing its motion after it has already left the ground.
Make sure that if you have Joules, you have kg, not grams.
K = 1/2 mv^2
10 = 1/2 (.2) v^2
10 = .1v^2
100=v^2
10 = v
So, the motion of the ball is
h = 10t - 4.9t^2
max height at t = -10/-9.8 = 1.02
h(1.02) = 5.10
The ball should come pretty close to this height, since we know its initial velocity. The only thing that could affect its height is air resistance. Any inelasticity of the rubber does not come into play, since we are only analyzing its motion after it has already left the ground.
4 answers