Find the angle of projection of a projectile for which its maximum height and horizontal range are equal. (Ans: 76°)



Given:

Maximum height h = horizontal range R



To Find:

Angle of projection = 𝞱 = ?


Solution:

For projectile motion, we have

Maximum height = h = vi² `\frac {sin² θ}{2g}` -----(1)


and


Horizontal range = R = vi² `\frac {sin (2θ)}{g}` -----(2)


As given {h = R} So, comparing Equation (1) and (2)


vi² `\frac {sin² θ}{2g}` = vi² `\frac {sin (2θ)}{g}`


or ( by simplifying)


`\frac {sin² θ}{2}` = sin 2θ


As Sin 2θ = 2 sinθ cosθ, so


`\frac {sin² θ}{2}` = 2 sinθ cosθ


or


`\frac {sin² θ}{sinθ cosθ}` = 2ｘ2


or


`\frac {sin θ}{cosθ}` = 4


or


tan θ = 4


or


θ = tan⁻¹ (4)


θ = 75.6° ≈ 76°  ---------------Ans.


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