Question

A body is projected with a velocity u so that the horizontal range is twice the maximum height. Then the maximum height is

Answer 1

H = (R/4) Tanθ

Given that R = 2H

So,
H = (H/2) Tanθ
Tanθ = 2

If Tanθ = 2 then
Sinθ = 2 / √5
Cosθ = 1 / √5

H = u² sin²θ/(2g)
= u² × (4/5) / (2g)
= 2u²/(5g)

∴ Maximum height is 2u²/(5g)

Answer 2

given

u= initial velcity of the projectile

θ= angle of projection.

According to question,

R=2H.

To find- maximum height

Step 1 R=2H

u^2sin2theta /g = u^2sin2theta/2gg

2sinθ.cosθ=sinθ.sinθ

tanθ=2

Step 2 :-

H=u^2sin^2 theta/2g

H =2u^2/5g

so,height is 2u^2/5g

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