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 u = 0 and
For second stone \(\frac{u^{2}}{2g} = 4h \Rightarrow u^{2} = 8gh\)
∴ ∴ \(u = \sqrt{2gh}\)
Now, \(h_{1} = \frac{1}{2} g t^{2}\)
\(h_{2} = \sqrt{8 g h t} - \frac{1}{2} g t^{2}\)
where, t =time to cross each other.
\(\boxed{ } \quad h_1 + h_2 = h\)
⇒ ⇒ \(\frac{1}{2}gt^{2} + \sqrt{8ght} - \frac{1}{2}gt^{2} = h\) ⇒ ⇒ \(t = \frac{h}{\sqrt{2gh}} = \sqrt{\frac{2h}{g}}\)
Prepare Smarter with CGP Edu
Get practice questions, solutions, and test series in one place.
Write a Review
Share your experience with this question and solution.
Commentary
Send your comment, doubt, correction, or feedback to admin.
Similar Questions
Explore conceptually related problems