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`        A ball is projected such that it's horizontal range is n times of the maximum height then find out the ratio of pe to ke at maximum height?`
7 months ago

Arun
23514 Points
```							Dear student u^2 sin 2theta /g = n u^2 sin^2 theta/2g 2 sin theta cos theta = n sin^2 theta /2 4 cos theta = n *sin theta tan theta = 4/n I think now you can calculate
```
7 months ago
Khimraj
3008 Points
```							A ball is projected in a manner such that its horizontal range is n times of the maximum height. Then find out the ratio of potential energy to kinetic energy at maximum height.​Let say Velocity  VVCosα = Horizontal SpeedVsinα = Vertical Speedat T max height reached so Vsinα becomes = 00 = Vsinα - gT=> T =  Vsinα/gVertical distance 0² - (Vsinα)² = 2(-g)S=> S = V²sin²α/2gTime to reach ground = 2T = 2Vsinα/gHorizontal Distance = Vcosα * 2Vsinα/g = 2V²CosαSinα/gHorizontal Distance = n * Vertical distance => 2V²CosαSinα/g = n * V²sin²α/2g=> 4Cosα =  nsinα=> sinα/Cosα = 4/nPotential Energy at Height = mgh =  mg V²sin²α/2g  = (1/2)mV²sin²αKinetic energy at max height = (1/2)m(VCosα)² = (1/2)mV²cos²αPotential Energy/Kinetic energy  = (1/2)mV²sin²α/ (1/2)mV²cos²α= (sinα/Cosα)²=  (4/n)²= 16/n²the ratio of potential energy to kinetic energy at maximum height.​ = 16/n²
```
7 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