Question

The horizontal range of a projectile is maximum when the angle of projection is ____.

• A. 0°
• B. 30°
• C. 45°
• D. 60°
• E. 90°

Hint:

At 45°, the projectile has the perfect balance between vertical height (to stay in the air longer) and horizontal velocity (to move forward faster). At angles like 90°, you get height but no forward distance!

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The horizontal range of a projectile is the total horizontal distance it travels before hitting the ground.
This distance depends on the initial launch velocity and the angle of projection.
Step 2: Key Formula or Approach:
The formula for the horizontal range $R$ of a projectile is $R = \frac{u^2 \sin(2\theta)}{g}$, where $u$ is initial velocity, $\theta$ is the angle of projection, and $g$ is acceleration due to gravity.
Step 3: Detailed Explanation:
For a given constant initial velocity $u$, the range $R$ is directly proportional to $\sin(2\theta)$.
To achieve the maximum range $R_{\text{max}}$, the value of $\sin(2\theta)$ must be at its maximum possible value.
The maximum value of the sine function is 1, which occurs when its argument is $90^{\circ}$.
Therefore, we set up the equation:
\[ \sin(2\theta) = 1 \]
\[ 2\theta = 90^{\circ} \]
Solving for the angle $\theta$:
\[ \theta = \frac{90^{\circ}}{2} = 45^{\circ} \]
Step 4: Final Answer:
The horizontal range of a projectile becomes maximum when the angle of projection is $45^{\circ}$.