(i) P.E. at its maximum height (ii) K.E. when it hits the ground

Question

A 500 g stone is thrown up with a velocity of $15 ms^{-1}$. Find its
(i) P.E. at its maximum height
(ii) K.E. when it hits the ground
(56.25 J, 56.25 J)

Accepted Answer

Solution:

Mass of stone = m = 500 g = $\frac{500}{1000}$ kg = 0.5 kg

Velocity = v = 15 $ms^{-1}$

Potential energy = P.E. = ?
Kinetic energy = K.E. = ?
Loss of K.E. = Gain in P.E.

$\frac{1}{2}m v_\_{f}^{2}$ – $\frac{1}{2}m v\_{i}^{2}$ = mgh

As the velocity of the stone at maximum height become zero, therefore, $v_{f}$ = 0

$\frac{1}{2} imes 0.5 imes (0) -\frac{1}{2} imes 0.5 imes (15)^{2}$ = mgh

$\frac{-1}{2} imes 0.5 imes 225$ = mgh

$-$56.25 = mgh

mgh = $-$56.25J

Since energy is always positive, therefore

P.E. = 56.25 J

K.E. = $\frac{1}{2}m v ^{2}$

K.E. = $\frac{1}{2} imes 0.5 imes (15)^{2}$

= $\frac{1}{2} imes 0.5 imes 225$

= 56.25 J