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A stone is dropped from a height of $100 \mathrm{~m}$, while another one is projected vertically upwards from the ground with a velocity of $25 \mathrm{~m} / \mathrm{s}$ at the same time.
The time in seconds after which they will have the same height is (acceleration due to gravity, $g=10 \mathrm{~ms}^{-2}$ )
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The time in seconds after which they will have the same height is (acceleration due to gravity, $g=10 \mathrm{~ms}^{-2}$ )
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Verified Answer
The correct answer is:
4
Let stones meet after time $t$ seconds.
Then distance travelled by first stone in $t$ seconds $=100-h(\mathrm{~m})$ and distance travelled by Ind stone in $t$ second is $h(\mathrm{~m})$. So, we have
For first stone, $\quad-(100-h)=\frac{-1}{2} g t^2$
For second stone,
$$
h=25 t-\frac{1}{2} g t^2
$$
Substituting for $\frac{1}{2} g t^2$ from Eq. (i) in Eq. (ii), we get
$$
\begin{aligned}
& h=25 t-(100-h) \Rightarrow 25 t=100 \\
& t=4 \mathrm{~s}
\end{aligned}
$$
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