Projectile Motion Range-to-Height Ratio – Rankers Physics

Ground to Ground Projectile: Practice Problem & Solution

A particle is projected with a speed of \( 20 \, \text{m s}^{-1} \) from level ground at an angle \( \theta \) equal to \( 45^\circ \) from horizontal. The ratio of maximum height attained by the body to horizontal range acquired by the body will be
\( \frac{2}{1} \)
\( \frac{1}{2} \)
\( \frac{1}{4} \)
\( \frac{3}{2} \)

Solution Explained:

To solve this problem, we apply the core principles of Ground to Ground Projectile. Understanding the underlying formula is key to arriving at the correct answer below:

The ratio of maximum height \( H \) to horizontal range \( R \) is given by \( \frac{H}{R} = \frac{\tan \theta}{4} \). For \( \theta = 45^\circ \), \( \tan 45^\circ = 1 \), leading to \( \frac{H}{R} = \frac{1}{4} \).

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