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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?
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1663 Upvotes
Verified Answer
The correct answer is:
$10 \mathrm{~s}$
$10 \mathrm{~s}$
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, $h=u t+\frac{1}{2} g t^2$
$$
\begin{aligned}
& 490=0+\frac{1}{2} \times 9.8 \times t^2 \\
& \Rightarrow \quad t^2=\frac{980}{9.8}=100 \\
& \therefore \quad t=\sqrt{100}=10 \mathrm{sec}
\end{aligned}
$$
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