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 find the maximum height reached by the projectile, we use the vertical component of the initial velocity and the formula for the maximum height in projectile motion. The vertical component of the initial velocity can be calculated using the sine function:

\( v_{0y} = v_0 \sin \theta = 20 \, \text{m/s} \times \sin 30^\circ \)

We know that \( \sin 30^\circ = 0.5 \), so:

\( v_{0y} = 20 \times 0.5 = 10 \, \text{m/s} \)

The formula for the maximum height \( h \) is given by:

\( h = \frac{{v_{0y}^2}}{2g} \)

Substitute \( v_{0y} = 10 \, \text{m/s} \) and \( g = 10 \, \text{m/s}^2 \):

\( h = \frac{{10^2}}{2 \times 10} = \frac{100}{20} = 5 \, \text{m} \)

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

Answer 2

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 \( 5 \, \text{m} \).