Question

A ball of mass $5\text{kg}$ is dropped from a cliff of height $75\text{m}$. What is the ratio of the kinetic energy to the potential energy of the ball when it reaches the halfway point of the fall?

• A. $\text{2}$
• B. $\text{1}$
• C. $\frac{1}{2}$
• D. $\frac{3}{4}$

Answer 1

The correct option is B $\text{1}$

According to the law of conservation of energy, the total energy of the ball remains constant.
$\therefore$ PE at the top of the cliff is equal to the sum of kinetic and potential energies half way through the fall. $(KE$ at top of the cliff $=0)$
Let $h=$ total height of the cliff
$h_{\text{half}}=$ Half of the total height of cliff

Then,
$(PE{)}_{\text{Top}}=K{E}_{\text{half}}+(PE{)}_{\text{half}}$
$\Rightarrow mgh=\frac{1}{2}m{v}_{\text{half}}^2+mg{h}_{\text{half}}$
$\Rightarrow 2gh=v_{\text{half}}^2+2g{h}_{\text{half}}$
Here $h=2h_{\text{half}}$
$\Rightarrow v_{\text{half}}^2=2g\left(2h_{\text{half}}\right)-2g{h}_{\text{half}}$
$\Rightarrow v_{\text{half}}=\sqrt{4g{h}_{\text{half}}-2g{h}_{\text{half}}}=\sqrt{2g{h}_{\text{half}}}$

The ratio of kinetic to potential energy half way through the fall
$=\frac{(KE{)}_{\text{half}}}{(PE{)}_{\text{half}}}=\frac{\frac{1}{2}m{v}_{\text{half}}^2}{mg{h}_{\text{half}}}=\frac{m\times {\left(\sqrt{2g{h}_{\text{half}}}\right)}^2}{2\times mg{h}_{\text{half}}}$
$\Rightarrow \frac{(KE{)}_{\text{half}}}{(PE{)}_{\text{half}}}=1$