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

For projectile motion, the maximum height \( H \) is given by the formula: \[ H = \frac{v_y^2}{2g} \] Where: - \( v_y = v \sin \theta \) is the vertical component of the initial velocity, - \( g = 9.8 \, \text{m/s}^2 \) is the acceleration due to gravity. Given: - \( v = 20 \, \text{m/s} \), - \( \theta = 30^\circ \). First, calculate \( v_y \): \[ v_y = 20 \sin 30^\circ = 20 \times 0.5 = 10 \, \text{m/s} \] Now, use the formula for the maximum height: \[ H = \frac{10^2}{2 \times 9.8} = \frac{100}{19.6} \approx 5.1 \, \text{m} \] Thus, the maximum height reached by the projectile is approximately \( 10 \, \text{m} \).