Question

A projectile is launched with an initial velocity of \( 20 \, \text{m/s} \) at an angle of \( 30^\circ \) with the horizontal. What is the maximum height reached by the projectile? (Take \( g = 10 \, \text{m/s}^2 \))

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

Hint:

Maximum height in projectile motion is \( H = \frac{u_y^2}{2g} \), where \( u_y \) is vertical velocity.

Answer 1

The Correct Option is A

The maximum height of the projectile is determined using its initial vertical velocity and the standard projectile motion formula for maximum height. The initial vertical velocity component is calculated via the sine function: \( v_{0y} = v_0 \sin \theta = 20 \, \text{m/s} \times \sin 30^\circ \). Given that \( \sin 30^\circ = 0.5 \), the initial vertical velocity is: \( v_{0y} = 20 \times 0.5 = 10 \, \text{m/s} \). The maximum height formula is \( h = \frac{{v_{0y}^2}}{2g} \). Substituting \( v_{0y} = 10 \, \text{m/s} \) and \( g = 10 \, \text{m/s}^2 \) yields: \( h = \frac{{10^2}}{2 \times 10} = \frac{100}{20} = 5 \, \text{m} \). Therefore, the projectile reaches a maximum height of \( 5 \, \text{m} \).