Question

A particle is dropped from a height $h$. Another particle which is initially at a horizontal distance $d$ from the first is simultaneously projected with a horizontal velocity $u$ and the two particles just collide on the ground. Then :

• A. $d^2=\frac{u^{2}h}{2g}$
• B. $d^2=\frac{2u^{2}h}{g}$
• C. $d^2=\frac{u^{2}h}{g}$
  
• D. $d^2=\frac{u^{2}h}{3g}$
  

Answer 1

The correct option is B $d^2=\frac{2u^{2}h}{g}$

Time for the 1st particle to reach the ground: $t=\sqrt{\frac{2h}{g}}$
The 2nd particle travels a distance d in time t at constant horizontal speed u.
$\therefore d=ut$
$d=u\sqrt{\frac{2h}{g}}$
$\Rightarrow d^2=\frac{2u^{2}h}{g}$
.