Question

A projectile is fired with an initial speed of \( 20 \, \text{m/s} \) at an angle of \( 30^\circ \) above the horizontal. Find the maximum height reached by the projectile.

• A. \( 10 \, \text{m} \)
• B. \( 20 \, \text{m} \)
• C. \( 5 \, \text{m} \)
• D. \( 15 \, \text{m} \)

Hint:

In projectile motion, the maximum height is determined by the vertical component of the initial velocity and the acceleration due to gravity. Always break the initial velocity into horizontal and vertical components.

Answer 1

The Correct Option is A

The maximum height \( H \) for projectile motion is determined by the formula \( H = \frac{v_y^2}{2g} \), where \( v_y \) represents the vertical component of the initial velocity and \( g \) is the acceleration due to gravity (\( 9.8 \, \text{m/s}^2 \)). Given an initial velocity \( v = 20 \, \text{m/s} \) and an angle \( \theta = 30^\circ \), the vertical component \( v_y \) is calculated as \( v_y = 20 \sin 30^\circ = 20 \times 0.5 = 10 \, \text{m/s} \). Subsequently, the maximum height is computed using the formula \( H = \frac{10^2}{2 \times 9.8} = \frac{100}{19.6} \approx 5.1 \, \text{m} \). Therefore, the projectile reaches an approximate maximum height of \( 5.1 \, \text{m} \).