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An object is thrown vertically upward with some speed. It crosses 2 points p,q which are seperated by h meter. If tp is the time between p  and highest point and coming back and tq is the time between q and highest point and coming back ,relate acceleration due to gravity,tp,tq, and h

5 months ago

Arun
23330 Points

Dear Sumedh

I have attached an image containing solution.

Regards

5 months ago
Khimraj
3008 Points

Let A be the highest point and h’ the distance between the point p and A. Here, tp = time taken in going from p to A and again from A to p A O q p h h' time taken in coming from A to p = p t 2 . Similary, time taken in falling from A to q = q t 2 . Now, h’ = 2 2 p p 1 t gt g 2 2 8     =   ....(i) and h’ – h = 2 2 q q 1 t gt g 2 2 8     =   ....(ii) Subtracting (ii) from (i), we get h = 2 2 p q 2 2 p q gt gt g (t t ) 8 8 8 − = − or, g = 2 2 p q 8h
5 months ago
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### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions