A stone is dropped from a height h. Simultaneously, another stone is thrown up from the ground which reaches a height 4 h. The two stones cross each other after time
Text Solution
Verified by ExpertsA
For first stone \(\oint f = 0\) and
For second stone \(\frac{u^2}{2g} - 4h \to u^2 - 8gh\)
\ \ \(u = \sqrt{2gh}\)
Now, \(h_1 - \frac{1}{2} g \vec{v}^2\)
\(h_2 - \sqrt{8ght} - \frac{1}{2}gt^2\)
where, t =time to cross each other.
\(\therefore \mathbf{h}_1 + \mathbf{h}_2 \quad \mathbf{h}\)
⇒ ⇒ \(\frac{1}{2}gt^{2} + \sqrt{8ght} - \frac{1}{2}gt^{2} - h\) ⇒ ⇒ \(\tau \quad \frac{\hbar}{\sqrt{8 g \pi}} \quad \psi \sqrt{\frac{\hbar}{8 g}}\)
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