A particle (a) is dropped from a height and another particle (b) is projected in the horizontal direction with a speed of $5 \mathrm{~ms}^{-1}$ from the same height, then the correct statement is:

• A

  particle (a) will reach at the ground first with respect to particle (b)
• B

  particle (b) will reach at the ground first with respect to particle (a)
• Correct answer
  C

  both particle will reach at the ground simultaneously
• D

  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 $\mathrm{S}=u t+\frac{1}{2}+a t^2$ $\Rightarrow h=\frac{1}{2} g+t_n^2 \Rightarrow t_{\mathrm{B}}=\sqrt{\frac{2 h}{g}}$

Hence, the correct answer is option: 'C'.