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

To determine the maximum height reached by the projectile, we use the formula for vertical motion. The vertical component of the initial velocity, \( V_{0y} \), is given by \( V_{0y} = V_0 \sin \theta \), where \( V_0 = 20 \, \text{m/s} \) and \( \theta = 30^\circ \).

First, we calculate \( V_{0y} \):

\( V_{0y} = 20 \, \text{m/s} \times \sin 30^\circ = 20 \, \text{m/s} \times 0.5 = 10 \, \text{m/s} \).

Next, using the formula for maximum height \( H \), \( H = \frac{V_{0y}^2}{2g} \), where \( g = 10 \, \text{m/s}^2 \), we find:

\( H = \frac{(10 \, \text{m/s})^2}{2 \times 10 \, \text{m/s}^2} = \frac{100 \, \text{m}^2/\text{s}^2}{20 \, \text{m/s}^2} = 5 \, \text{m} \).

Thus, the maximum height reached by the projectile is \( 5 \, \text{m} \).