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A boy standing on a moving truck throws a projectile such that he is able to catch it back after the truck has moved $100 \mathrm{~m}$. If the truck is moving horizontally along a straight line with a constant speed $30 \mathrm{~m} / \mathrm{s}$, at what speed (relative to the truck) must the projectile is thrown. (Assume, $g=10 \mathrm{~m} / \mathrm{s}^2$ )
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Verified Answer
The correct answer is:
$\frac{50}{3} \mathrm{~m} / \mathrm{s}$
Given, velocity of truck, $v=30 \mathrm{~m} / \mathrm{s}$ time taken by truck to move a distance of $100 \mathrm{~m}$,
$$
t=\frac{100}{30} \Rightarrow t=\frac{10}{3} \mathrm{~s}
$$
If $v_1$ be the velocity of projectile in upward direction then time taken by the projectile to reach at maximum height is $t_1$, then
$$
\begin{aligned}
0 & =v_1-g t, \\
v_1 & =g t_1
\end{aligned}
$$
But, $\quad t_1=\frac{t}{2}=\frac{\frac{10}{3}}{2}=\frac{5}{3} \mathrm{~s}$
$$
\therefore \quad v_1=g \times \frac{5}{3}=\frac{10 \times 5}{3}=\frac{50}{3} \mathrm{~m} / \mathrm{s}
$$
$$
t=\frac{100}{30} \Rightarrow t=\frac{10}{3} \mathrm{~s}
$$
If $v_1$ be the velocity of projectile in upward direction then time taken by the projectile to reach at maximum height is $t_1$, then
$$
\begin{aligned}
0 & =v_1-g t, \\
v_1 & =g t_1
\end{aligned}
$$
But, $\quad t_1=\frac{t}{2}=\frac{\frac{10}{3}}{2}=\frac{5}{3} \mathrm{~s}$
$$
\therefore \quad v_1=g \times \frac{5}{3}=\frac{10 \times 5}{3}=\frac{50}{3} \mathrm{~m} / \mathrm{s}
$$
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