Question 30 of 130

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$5 \mathrm{~s}$C$10 \mathrm{~s}$D$\sqrt{10} \mathrm{~s}$

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Answer: (C) $10 \mathrm{~s}$

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, $h=u t+\frac{1}{2} g t^2$
$$\begin{aligned} & 490=0+\frac{1}{2} \times 9.8 \times t^2 \\ & \Rightarrow \quad t^2=\frac{980}{9.8}=100 \\ & \therefore \quad t=\sqrt{100}=10 \mathrm{sec} \end{aligned}$$