A particle (a) is dropped from a height and another particle (b) is projected in the horizontal direction…
- particle (a) will reach at the ground first with respect to particle (b)
- particle (b) will reach at the ground first with respect to particle (a)
- both particle will reach at the ground simultaneously
- both particles will reach at the ground with the same speed
Solution

For particle A
$\begin{aligned}
& h=\frac{1}{2} g t^2 \\
& t_{\mathrm{A}}=\sqrt{\frac{2 h}{g}}
\end{aligned}$
For particle B. In vertical direction
Using $\mathrm{S}=u t+\frac{1}{2}+a t^2$
$\Rightarrow h=\frac{1}{2} g+t_n^2 \Rightarrow t_{\mathrm{B}}=\sqrt{\frac{2 h}{g}}$
Asked in: NEET 2002
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