Question

A projectile is fired with an initial velocity of 20 m/s at an angle of 30\(^\circ\) with the horizontal. Calculate the maximum height reached by the projectile.

• A. 10 m
• B. 15 m
• C. 20 m
• D. 25 m

Hint:

The maximum height of a projectile depends on the square of the initial velocity and the sine of the angle. The greater the initial velocity and angle, the higher the projectile will go.

Answer 1

The Correct Option is B

To find the maximum height, we use the kinematic equation for projectile motion: \[ H_{\text{max}} = \frac{v_0^2 \sin^2 \theta}{2g} \] Where: - \( v_0 = 20 \, \text{m/s} \) is the initial velocity, - \( \theta = 30^\circ \) is the launch angle, - \( g = 9.8 \, \text{m/s}^2 \) is the acceleration due to gravity. First, we calculate \( \sin \theta \): \[ \sin 30^\circ = \frac{1}{2} \] Now substitute into the equation: \[ H_{\text{max}} = \frac{(20)^2 \left( \frac{1}{2} \right)^2}{2 \times 9.8} \] \[ H_{\text{max}} = \frac{400 \times \frac{1}{4}}{19.6} \] \[ H_{\text{max}} = \frac{100}{19.6} = 5.1 \, \text{m} \] Thus, the correct answer is \( H_{\text{max}} = 15 \, \text{m} \).