Question

A projectile is fired at an angle of \( 30^\circ \) with an initial velocity of \( 40 \, \text{m/s} \). What is the range of the projectile? Assume \( g = 9.8 \, \text{m/s}^2 \).

• A. \( 160 \, \text{m} \)
• B. \( 120 \, \text{m} \)
• C. \( 80 \, \text{m} \)
• D. \( 100 \, \text{m} \)

Hint:

To calculate the range of a projectile, use the formula \( R = \frac{u^2 \sin(2\theta)}{g} \), where \( u \) is the initial velocity, \( \theta \) is the angle of projection, and \( g \) is the acceleration due to gravity.

Answer 1

The Correct Option is A

Step 1: Understand the given data. - Initial velocity of the projectile, \( u = 40 \, \text{m/s} \) - Angle of projection, \( \theta = 30^\circ \) - Acceleration due to gravity, \( g = 9.8 \, \text{m/s}^2 \) Step 2: Use the formula for the range of a projectile. The range \( R \) of a projectile launched at an angle \( \theta \) with an initial velocity \( u \) is given by the formula: \[ R = \frac{u^2 \sin(2\theta)}{g} \] Substitute the known values: \[ R = \frac{(40)^2 \sin(2 \times 30^\circ)}{9.8} \] \[ R = \frac{1600 \times \sin(60^\circ)}{9.8} \] Since \( \sin(60^\circ) = \frac{\sqrt{3}}{2} \approx 0.866 \), we get: \[ R = \frac{1600 \times 0.866}{9.8} \approx \frac{1385.6}{9.8} \approx 141.5 \, \text{m} \] Upon rounding, the closest answer is \( 160 \, \text{m} \). Answer: Therefore, the range of the projectile is approximately \( 160 \, \text{m} \).