Question

A stone projected with a velocity u at an angle with the horizontal reaches maximum height h1. When it is projected with velocity u at an angle 2 with the horizontal, it reaches maximum height h2. The relation between the horizontal range r of the projectile, h1 and h2 is

Answer 1

Let the velocity of projection of the stone be uwith angle of projection θ with the horizontal direction.
The vertical component of the velocity of projection is usinθ and the horizontal component is ucosθ

Now if the time of flight be T then the object will return to the ground after T sec and during this T sec its total vertical displacement h will be zero. So applying the equation of motion under gravity we can write

h=usinθ×T+12gT2

⇒0=usinθT−12×g×T2 where g=acceleration due to gravity

∴T=2usinθg
The horizontal displacement or range during this T sec is

R=ucosθ×T

⇒R=u2sin(2θ)g.......[1]

If H be the maximum height attained by the stone we can write

02=u2sin2θ−2gH

⇒H=u2sin2θ2g....[2]

CALCULATION USING ABOVE RELATIONS

For 1st case

when θ=α,H=h1

So horizontal range

R1=u2sin(2α)g.....[3]

and h1=u2sin2α2g....[4]

For 2nd case

when θ=90−α,H=h2

So horizontal range

R2=u2sin(2(90−α))g

⇒R2=u2sin(180−2α)g

⇒R2=u2sin(2α)g....[5]

and h2=u2sin2(90−α)2g

⇒h2=u2cos2(α)2g....[6]

In both the cases the range is same.So let R1=R2=R

Hence range

R=u2sin(2α)g

Now multiplying [4] and [6] we get

h1h2=u4sin2αcos2α(2g)2

⇒h1h2=(u2×2sinαcosα)2(4g)2

⇒h1h2=(u2×sin2α)2(4g)2

⇒√h1h2=u2×sin2α4g

⇒√h1h2=R4

⇒R=4√h1h2