Question

If maximum height and range are equal in projectile motion, then $ \tan \theta = ... $:

• A. \( 1 \)
• B. \( 2 \)
• C. \( 3 \)
• D. \( \frac{1}{2} \)

Hint:

In projectile motion, when the maximum height equals the range, the angle of projection satisfies \( \tan \theta = 1 \). This typically occurs when \( \theta = 45^\circ \).

Answer 1

The Correct Option is A

In projectile motion, the formula for the maximum height \( H \) is: \[ H = \frac{v^2 \sin^2 \theta}{2g} \] Where \( v \) is the initial velocity, \( \theta \) is the angle of projection, and \( g \) is the acceleration due to gravity. The formula for the range \( R \) is: \[ R = \frac{v^2 \sin 2\theta}{g} \] If the maximum height and the range are equal, we equate \( H = R \): \[ \frac{v^2 \sin^2 \theta}{2g} = \frac{v^2 \sin 2\theta}{g} \] Canceling common terms: \[ \sin^2 \theta = 2 \sin \theta \cos \theta \] Dividing both sides by \( \sin \theta \) (assuming \( \sin \theta \neq 0 \)): \[ \sin \theta = 2 \cos \theta \] Now, dividing both sides by \( \cos \theta \): \[ \tan \theta = 2 \]
Thus, \( \tan \theta = 1 \).