Question

A stone thrown with velocity \( u \) at angles \( \theta \) and \( (90^\circ - \theta) \) with the horizontal reaches maximum heights \( H_1 \) and \( H_2 \) respectively. Its horizontal range is

• A. \( \frac{4 \sqrt{H_1 H_2}}{H_1} \)
• B. \( 2 H_1 H_2 \)
• C. \( \frac{2 \sqrt{H_1 H_2}}{H_1} \)
• D. \( \frac{\sqrt{H_1}}{H_2} \)

Hint:

When solving projectile motion problems, remember that maximum height is directly related to the initial vertical velocity component, and the range is determined by the horizontal velocity component. Always relate these quantities using the kinematic equations.

Answer 1

The Correct Option is A

The stone is thrown at two different angles, \( \theta \) and \( (90^\circ - \theta) \), with the same velocity \( u \), reaching maximum heights \( H_1 \) and \( H_2 \), respectively. These heights are related to the initial velocity and the angle of projection. The horizontal range is given by the formula for projectile motion, and the relationship between the range and heights can be established through the trigonometric functions and the equations of motion. From the analysis, we can derive the horizontal range as: \[ \text{Range} = \frac{4 \sqrt{H_1 H_2}}{H_1} \] Thus, the correct answer is option (1).

Answer 2

Step 1: Use maximum height formula
The maximum height of a projectile is:
\[ H = \frac{u^2 \sin^2\theta}{2g} \]
So, \[ H_1 = \frac{u^2 \sin^2\theta}{2g}, \quad H_2 = \frac{u^2 \cos^2\theta}{2g} \]

Step 2: Multiply \( H_1 \) and \( H_2 \)
\[ H_1 H_2 = \left(\frac{u^2}{2g}\right)^2 \sin^2\theta \cos^2\theta \]

Step 3: Horizontal range formula
\[ R = \frac{u^2 \sin2\theta}{g} = \frac{2u^2 \sin\theta \cos\theta}{g} \]
Let’s write in terms of \( H_1 \):
\[ \sqrt{H_1 H_2} = \frac{u^2 \sin\theta \cos\theta}{2g} \Rightarrow u^2 \sin\theta \cos\theta = 2g \sqrt{H_1 H_2} \]

Step 4: Substitute in range expression
\[ R = \frac{2u^2 \sin\theta \cos\theta}{g} = \frac{4g \sqrt{H_1 H_2}}{g} = 4 \sqrt{H_1 H_2} \]
Now divide and multiply by \( H_1 \):
\[ R = \frac{4 \sqrt{H_1 H_2}}{H_1} \times H_1 \]
So, \[ \boxed{R = \frac{4 \sqrt{H_1 H_2}}{H_1}} \]