Question

A 0.5 kg ball is thrown up into the air with an initial speed of 6 m/s . at what height does the gravitational potential energy of the ball equal its initial kinetic energy?

Answer 1

Let's assume that the height from which the ball was thrown is 0 meters. Because energy is conserved, the initial energy is equal to the final energy:

$PE_i + KE_i = PE_f + KE_f$

The ball was thrown with an initial velocity, and we're assuming that the height from which it was thrown is 0, so initially, gravitational potential energy is 0 and kinetic energy is nonzero. At the peak of the ball's motion, velocity is going to be 0, but the height is nonzero, so the kinetic energy at the peak is going to be zero, but the gravitation potential energy is going to be nonzero. Because of the law of conservation of energy, we know then that the initial kinetic energy is going to be equal to the gravitational potential energy at the peak of the ball's motion:

$KE=PE \\ 0.5mv^{2}=mgh \\ 0.5v^{2} =gh \\ h=\frac{0.5v^{2}}{g} \\ h= \frac{0.5(6)^{2}}{9.8} \\ h=1.8$

So the ball's gravitational potential energy is going to equal its initial kinetic energy at a height of 1.8 meters.