Angle of Projection – Rankers Physics
Topic: Kinematics
Subtopic: Ground to Ground Projectile

Angle of Projection

At what angle to the horizontal should an object be projected so that the maximum height reached is equal to half of the horizontal range?
\(\tan^{-1} (1)\)
\(\tan^{-1} (2)\)
\(\tan^{-1} (3)\)
\(\tan^{-1} (4)\)

Solution:

We require \(H = \frac{R}{2}\). Using the formulae \(H = \frac{u^2 \sin^2 \theta}{2g}\) and \(R = \frac{u^2 \sin 2\theta}{g}\), we get \(\frac{u^2 \sin^2 \theta}{2g} = \frac{u^2 \sin\theta\cos\theta}{g} ⇒ \tan\theta = 2 ⇒ \theta = \tan^{-1}(2)\).

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