A projectile moving vertically upwards with a velocity of $200 \mathrm{~ms}^{-1}$ breaks into two equal parts at a height of $490 \mathrm{~m}$. One part starts moving vertically upwards with a velocity of $400 \mathrm{~ms}^{-1}$. How much time it will take, after the break up with the other part to hit the ground?

• A

  $2 \sqrt{10} \mathrm{~s}$
• B

  $5 \mathrm{~s}$
• Correct answer
  C

  $10 \mathrm{~s}$
• D

  $\sqrt{10} \mathrm{~s}$

Solution

Momentum before explosion = Momentum after explosion $$ \begin{gathered} m \times 200 \hat{j}=\frac{m}{2} \times 400 \hat{j}+\frac{m}{2} v \\ =\frac{m}{2}(400 \hat{j}+v) \\ \Rightarrow \quad 400 \hat{j}-400 \hat{j}=v \\ \therefore \quad v=0 \end{gathered} $$ i.e., the velocity of the other part of the mass, $v=0$ Let time taken to reach the earth by this part be $t$ Applying formula, $h=u t+\frac{1}{2} g t^2$ $$ \begin{aligned} & 490=0+\frac{1}{2} \times 9.8 \times t^2 \\ & \Rightarrow \quad t^2=\frac{980}{9.8}=100 \\ & \therefore \quad t=\sqrt{100}=10 \mathrm{sec} \end{aligned} $$

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