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A body is projected horizontally from the top of a tall tower with a velocity of $30 \mathrm{~ms}^{-1}$. At time $t_1$, its horizontal and vertical components of the velocity are equal and at time $t_2$, its horizontal and vertical displacements are equal. Then $t_2-t_1$ is (take, $g=10 \mathrm{~ms}^{-2}$ )
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Verified Answer
The correct answer is:
$3 \mathrm{~s}$
As per first condition,
$$
\begin{gathered}
30=10 t_1 \\
t_1=3 \mathrm{~s}
\end{gathered}
$$
As per second condition,
$$
\begin{aligned}
30 t_2 & =\frac{1}{2} \times 10 t_2^2=t_2=6 \mathrm{~s} \\
\therefore \quad t_2-t_1 & =3 \mathrm{~s}
\end{aligned}
$$
$$
\begin{gathered}
30=10 t_1 \\
t_1=3 \mathrm{~s}
\end{gathered}
$$
As per second condition,
$$
\begin{aligned}
30 t_2 & =\frac{1}{2} \times 10 t_2^2=t_2=6 \mathrm{~s} \\
\therefore \quad t_2-t_1 & =3 \mathrm{~s}
\end{aligned}
$$
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