A particle is projected vertically up and another is let fall to meet at the same instant. If they have velocities equal in magnitude when they meet, the distances travelled by them are in the ratio of

For the projected body v₁ = u – gt ........ (1)

For the dropped body v₂ = gt ........... (2)

equating 1 & 2 t = u/2g

Dropped body projected body
$\begin{array}{l}{\mathrm{s}}_1=\frac{1}{2}{\mathrm{gt}}^2;\\ {\mathrm{s}}_1=\frac{1}{2}\cdot \mathrm{g}\frac{{\mathrm{u}}^2}{4{\mathrm{g}}^2}=\frac{{\mathrm{u}}^2}{8\mathrm{g}}\end{array}$

$\begin{array}{c}{\mathrm{s}}_2=\mathrm{ut}-\frac{1}{2}{\mathrm{gt}}^2;\mathrm{t}=\frac{\mathrm{u}}{2\mathrm{g}}\\ {\mathrm{s}}_2=\frac{{\mathrm{u}}^2}{2\mathrm{g}}-\frac{{\mathrm{u}}^2}{8\mathrm{g}}=\frac{3{\mathrm{u}}^2}{8\mathrm{g}}\end{array}$
Ratio $\frac{{\mathrm{s}}_1}{{\mathrm{s}}_2}=\frac{3}{1}$