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`        Question in image from physics kinematics.Solve it pls`
one year ago

Arun
23006 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   RegardsArun (askIITians forum expert)
```
one year 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