Question

Find angle of projectile for which the horizontal range and max height are equal.



Class - 11
Subject - Physics
Chapter -4​

Answer 1

Given:-

• For a projectile, it's horizontal range and maximum height is same

To find:-

• Angle of projection of the projectile

Answer:-

We have to use the formula,

▪R = [u² sin(2θ)] / [g]

▪H = [u² sin²(θ)] / [2g]

where R is the horizontal range, H is the maximum height, u is the initial velocity of projection, θ is the angle of projection, and g is the acceleration due to gravity.

According to the question, horizontal range and maximum height of the projectile is equal.

R = H

→ [u² sin(2θ)] / [g] = [u² sin²(θ)] / [2g]

→ [sin(2θ)] / [g] = [sin²(θ)] / [2g]

→ sin(2θ) = [sin²(θ)] / [2]

Now using the formula,

▪sin(2θ) = 2 sin(θ) cos(θ),

→ 2 sin(θ) cos (θ) = [sin²(θ)] / [2]

→ 2 sin(θ) cos (θ) = [sin(θ) sin(θ)] / [2]

→ 2 cos(θ) = [sin(θ)] / [2]

→ 4 cos(θ) = sin(θ)

→ [sin(θ)] / [cos(θ)] = 4

→ tan(θ) = 4

→ θ = tan⁻¹(4) Ans.