KCET · Physics · Motion In One Dimension
Two stones begin to fall from rest from the same height, with the second stone starting to fall ' \(\Delta \mathrm{t}\) ' seconds after the first falls from rest. The distance of separation between the two stones becomes ' \(H\) ', ' \(t_0\) ' seconds after the first stone starts its motion. Then \(t_0\) is equal to
- A \(\frac{\mathrm{H}}{\Delta \mathrm{t}}+\frac{\Delta \mathrm{t}}{2 \mathrm{~g}}\)
- B \(\frac{H}{g \Delta t}-\frac{\Delta t}{2}\)
- C \(\frac{H}{g \Delta t}+\frac{\Delta t}{2}\)
- D \(\frac{H}{g \Delta t}\)
Answer & Solution
Correct Answer
(C) \(\frac{H}{g \Delta t}+\frac{\Delta t}{2}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \mathrm{S}_1=\frac{1}{2} \mathrm{~gt}{ }^2 \\ & \mathrm{~S}_2=\frac{1}{2} \mathrm{~g}\left(\mathrm{t}_0-4 \mathrm{t}\right)^2 \\ & \mathrm{H}=\mathrm{S}_1-\mathrm{S}_2 \\ & =\frac{1}{2} \mathrm{gt}_0{ }^2-\frac{1}{2} \mathrm{~g}\left(\mathrm{t}_0-4 \mathrm{t}\right)^2\end{aligned}\)
\(\begin{aligned} & \mathrm{H}=\mathrm{gt}_0 4 \mathrm{t}-\frac{1}{2} \mathrm{~g} 4 \mathrm{t}^2 \\ & \mathrm{t}_0=\frac{\mathrm{H}}{\mathrm{g} \Delta \mathrm{t}}=\frac{1}{2} \Delta \mathrm{t}\end{aligned}\)
\(\begin{aligned} & \mathrm{H}=\mathrm{gt}_0 4 \mathrm{t}-\frac{1}{2} \mathrm{~g} 4 \mathrm{t}^2 \\ & \mathrm{t}_0=\frac{\mathrm{H}}{\mathrm{g} \Delta \mathrm{t}}=\frac{1}{2} \Delta \mathrm{t}\end{aligned}\)
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