Question

The range of projectile is maximum when the angle of projection is

• A. \( 30^\circ \)
• B. \( 45^\circ \)
• C. \( 60^\circ \)
• D. \( 90^\circ \)

Hint:

Launching at exactly \( 45^\circ \) provides the perfect balance between the vertical velocity component (which keeps the projectile in the air longer) and the horizontal velocity component (which moves the projectile forward faster).

Answer 1

The Correct Option is B

Concept: A projectile is launched from flat ground into a vacuum with an initial velocity magnitude \( v_0 \) at an inclination angle \( \theta \) relative to the horizontal. The total horizontal path distance traveled before returning to the launch height is defined as the horizontal range \( R \), given by the kinematic formula: \[ R = \frac{v_0^2\sin(2\theta)}{g} \] where \( g \) represents the local constant acceleration due to gravity. To find the maximum possible range for a fixed launch speed \( v_0 \), we maximize the value of the trigonometric sine function.

Step 1: Setting up the mathematical maximization condition.
In our horizontal range equation: \[ R = \left(\frac{v_0^2}{g}\right) \cdot \sin(2\theta) \] The initial velocity \( v_0 \) and the gravitational constant \( g \) are fixed positive scalar constants. Therefore, the horizontal range \( R \) reaches its maximum value when the variable trigonometric factor \( \sin(2\theta) \) reaches its highest possible mathematical value: \[ \sin(2\theta) = \text{Maximum value} \]

Step 2: Finding the maximum value of the sine function.
The mathematical sine function for any real angle is bounded between \(-1\) and \(+1\). Its maximum value is: \[ \max(\sin\phi) = 1 \] Therefore, we set our sine term equal to 1: \[ \sin(2\theta) = 1 \]

Step 3: Solving for the launch angle \( \theta \).
We find the principal angle where the sine function equals 1: \[ 2\theta = \sin^{-1}(1) \quad \Rightarrow \quad 2\theta = 90^\circ \] Dividing both sides by 2 isolates the target launch angle: \[ \theta = \frac{90^\circ}{2} = 45^\circ \] Thus, the horizontal range of a projectile is maximized when it is launched at an angle of exactly \( 45^\circ \). This corresponds to option (B).