Question

The ratio of maximum vertical to maximum horizontal distances travelled by a projectile with \( \theta \) as the angle of initial velocity with ground depends on

• A. \( \sin \theta \)
• B. \( \cos \theta \)
• C. \( \sin 2\theta \)
• D. \( \tan \theta \)

Hint:

The ratio of vertical to horizontal distance in projectile motion depends on \( \tan \theta \), where \( \theta \) is the angle of projection.

Answer 1

The Correct Option is D

Step 1: Analyze the motion of a projectile.
For projectile motion, the maximum vertical distance (height) is determined by the initial velocity and the angle of projection, while the maximum horizontal distance (range) depends on the initial velocity and the angle.
Step 2: Formula for vertical and horizontal distances.
The maximum vertical distance \( H \) is given by: \[ H = \frac{u^2 \sin^2 \theta}{2g}, \] and the maximum horizontal distance (range) \( R \) is given by: \[ R = \frac{u^2 \sin 2\theta}{g}. \]
Step 3: Find the ratio.
The ratio of maximum vertical to maximum horizontal distances is: \[ \frac{H}{R} = \frac{\frac{u^2 \sin^2 \theta}{2g}}{\frac{u^2 \sin 2\theta}{g}} = \frac{\sin^2 \theta}{2 \sin 2\theta}. \] Since \( \sin 2\theta = 2 \sin \theta \cos \theta \), the ratio simplifies to \( \frac{\sin \theta}{2 \cos \theta} = \tan \theta \).
Step 4: Conclusion.
Thus, the ratio of maximum vertical to maximum horizontal distances depends on \( \tan \theta \), which corresponds to option (D).