Question

estimate we be angle of projectile of a particle for which range R and max. hight H are equal

Answer 1

At Max height, vertical component of velocity becomes zero.

Use the Equation (Vf)^2 = (Vi)^2 - 2gS Where Vf = final velocity in vertical direction, Vi = Intial velocity in vertical direction and S is the Max height.

At Max height, Vf = 0, Vi= V sinθ

So we get

0 = (V sinθ)^2 - 2gS

or S = (V sinθ)^2/ 2g ——- Eq 1

Now

Range = V^2 Sin2θ/g ——- Eq 2

If you equate Eq 1 and 2 That is “Max Height = Range” and simplify, you’ll get

Tanθ = 4

Theta is about 76 degree

Answer 2

At Max height, vertical component of velocity becomes zero.

Use the Equation (Vf)^2 = (Vi)^2 - 2gS Where Vf = final velocity in vertical direction, Vi = Intial velocity in vertical direction and S is the Max height.

At Max height, Vf = 0, Vi= V sinθ

So we get

0 = (V sinθ)^2 - 2gS

or S = (V sinθ)^2/ 2g ——- Eq 1

Now

Range = V^2 Sin2θ/g ——- Eq 2

If you equate Eq 1 and 2 That is “Max Height = Range” and simplify, you’ll get

Tanθ = 4

Theta is about 76 degree

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