Question

1.A ball is projected in a manner such that itshorizontal range is n times of the maximumheight. Then find out the ratio of potentialenergy to kinetic energy at maximum height.​

Answer 1

Answer:

16/n²

Explanation:

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