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A ball released from the top of a tower travels 11/36 of the height of the tower in the last second of its journey. the height of the tower is?
3 years ago

Vikas TU
9762 Points

Assuming height of the tower h meter and time t we get,
=>h = 1/2gt^2 -------------------(i)
Distance travelled in (t-1) secs,
=>s = 1/2g(t-1)^2 ---------------(ii)
Subtract both the eqns. we get,
=>11h/36 = 1/2g(2t - 1) ------------(iii)
11/36 x 1/2gt^2 = 1/2g(2t-1)
=>11t^2 = 72t - 36
=>11t^2 - 72t + 36 = 0
On sollving further,
=>t = [72+/-60]/22
=>t = 6 sec OR 0.55 sec
Therefore.
h = 176.40 m
3 years 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