A 0.5kg ball thrown upwards hs an initial kinetic energy of 80J. What is the potential energy at max height?

Question

what is its max height?
What are its kinetic and potential energies when it has traveled 3/4 of the distance to its maximun height?
What is its velocity at this point?

Answer 1

KE = .5mv^2 = U = mgh
h = mv^2/2g, which turns out to be 80J/9.8m/s^2
or
8.16 m.
.75h = 6.12 m = h'
mgh' + .5mv'^2 = 80J (by conservation of energy)
simplify to m(gh' + .5v'^2) = 80J
By algebraic manipulation you wind up with √[(80J/m - gh')/.5]
or
14.1 m/s
KE = .5mv^2
U = mgh
Solving for those at that point should be simple.
Hope this helps.

Answer 2

Since neither the size nor the density (from which the size could be derived) of the ball are given, we can assume air resistance is negligible and to be ignored; i.e., the system is lossless. It's also implied that the ball is thrown straight up.

At maximum height, all the kinetic energy is converted to potential, so, 80J potential energy at max neight.

At the surface of the earth, we can use the approximation E=mgh (that's the small g) to find the max height. So the potential energy at 3/4 the max height is 3/4 the energy. That leaves 1/4 of the energy for kinetic energy, from which you can find the velocity by E = 1/2 mv^2 .

All this without even needing to find the maximum height.

Answer 3

These problems are simple to do and require very little calculation, if you remember that kinetic energy at bottom = potential energy at top: m*g*h = 10 0.2*9.8*h = 10 h = 10/1.96 h = 5.1m

Answer 4

use the formuals KE=(1/2)mv^2, PE=mgh=wh
The KE formula will give you its terminal velocity, then use a law of motion formula to find its height (or max height). Then plus its mass, gravity, and height into the PE formula. Have fun =)