A ball is falling freely from a height. When it reaches $10 \mathrm{~m}$ height from the ground its velocity…
A ball is falling freely from a height. When it reaches $10 \mathrm{~m}$ height from the ground its velocity is $v_0$. It collides with the ground and loses $50 \%$ of its energy and rises back to height of $10 \mathrm{~m}$. Then the velocity $v_0$ is
$7 \mathrm{~m} / \mathrm{s}$
$10 \mathrm{~m} / \mathrm{s}$
$14 \mathrm{~m} / \mathrm{s}$
$16 \mathrm{~m} / \mathrm{s}$
Solution
At height $10 \mathrm{~m}$, the kinetic energy is equal to potential energy then
$\frac{1}{2} m v_0^2=m g h$
$\frac{1}{2} v_0^2=9.8 \times 10$
$v_0^2=2 \times 9.8 \times 10$
$v_0=\sqrt{196}$
$v_0=14 \mathrm{~m} / \mathrm{s}$
