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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?
10 months ago

Arun
24489 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
10 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  V
VCosα = Horizontal Speed
Vsinα = Vertical Speed
at T max height reached so Vsinα becomes = 0
0 = Vsinα - gT
=> T =  Vsinα/g
Vertical distance
0² - (Vsinα)² = 2(-g)S
=> S = V²sin²α/2g
Time to reach ground = 2T = 2Vsinα/g
Horizontal Distance = Vcosα * 2Vsinα/g = 2V²CosαSinα/g
Horizontal Distance = n * Vertical distance
=> 2V²CosαSinα/g = n * V²sin²α/2g
=> 4Cosα =  nsinα
=> sinα/Cosα = 4/n

Potential Energy at Height = mgh =  mg V²sin²α/2g  = (1/2)mV²sin²α
Kinetic energy at max height = (1/2)m(VCosα)² = (1/2)mV²cos²α
Potential Energy/Kinetic energy  = (1/2)mV²sin²α/ (1/2)mV²cos²α
= (sinα/Cosα)²
=  (4/n)²
= 16/n²
the ratio of potential energy to kinetic energy at maximum height.​ = 16/n²
10 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