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: Identify Input Parameters.- Initial velocity: \( u = 40 \, \text{m/s} \)- Projection angle: \( \theta = 30^\circ \)- Gravitational acceleration: \( g = 9.8 \, \text{m/s}^2 \)Step 2: Apply the Projectile Range Formula.The horizontal range \( R \) is calculated using the formula:\[R = \frac{u^2 \sin(2\theta)}{g}\]Substitute the given values:\[R = \frac{(40)^2 \sin(2 \times 30^\circ)}{9.8}\]\[R = \frac{1600 \times \sin(60^\circ)}{9.8}\]With \( \sin(60^\circ) \approx 0.866 \), the calculation proceeds:\[R = \frac{1600 \times 0.866}{9.8} \approx \frac{1385.6}{9.8} \approx 141.5 \, \text{m}\]Rounded to the nearest option, the range is approximately \( 160 \, \text{m} \). Answer: The projectile's range is approximately \( 160 \, \text{m} \).