Question

A freely falling body crosses points $P,Q$ and $R$ with velocities $v,2v$ and $3v$ respectively. Find the ratio of the distances $PQ$ to $QR$.

Answer 1

In free fall,

$u=0$. Let $A$ be the initial point from where it is dropped.

So using third equation of motion.Where $H$ is displacement and $g$ is acceleration due to gravity

$v^2=2gH$

$v^2=2g\left(AP\right)$..(1)

$4v^2=2g\left(AQ\right)$..(2)

$9v^2=2g\left(AR\right)$...(3)

Substract (1) from (2)

$3v^2=2g(AQ-AP)=2g\left(PQ\right)$..(4)

Substract (2) from (3),

$5v^2=2g(AR-AQ)=2g\left(QR\right)$..(5)

Divide (4) and (5)

$\frac{3v^2}{5v^2}=\frac{PQ}{QR}$

$\frac{PQ}{QR}=\frac{3}{5}$