Question

A particle is projected with velocity \( u \) so that its horizontal range is three times the maximum height attained by it. The horizontal range of the projectile is given as \( \frac{nu^2}{25g} \), where value of \( n \) is : (Given 'g' is the acceleration due to gravity).

• A. 6
• B. 18
• C. 12
• D. 24

Hint:

Use the formulas for the horizontal range and maximum height of a projectile. Set up the given condition relating the range and maximum height to find the angle of projection. Once the angle is known (or trigonometric ratios of the angle are known), substitute these values back into the formula for the range to express it in the required form and find the value of \( n \).

Answer 1

The Correct Option is D

To solve the problem, we need to find the value of \( n \) for which the horizontal range \( R \) is three times the maximum height \( H \) of the projectile. The range is given as \( \frac{nu^2}{25g} \).

1. First, let's recall the formulas for the horizontal range and maximum height of a projectile:
   • The horizontal range \( R \) of a projectile is given by: \(R = \frac{u^2 \sin(2\theta)}{g}\)
   • The maximum height \( H \) attained by the projectile is: \(H = \frac{u^2 \sin^2(\theta)}{2g}\)
2. According to the problem, the horizontal range is three times the maximum height: \(R = 3H\)
3. Substitute the expressions for \( R \) and \( H \) in the equation: \(\frac{u^2 \sin(2\theta)}{g} = 3 \times \frac{u^2 \sin^2(\theta)}{2g}\)
4. Simplify the equation: \(\sin(2\theta) = \frac{3}{2} \sin^2(\theta)\)
5. Use the identity \(\sin(2\theta) = 2\sin(\theta)\cos(\theta)\): \(2\sin(\theta)\cos(\theta) = \frac{3}{2} \sin^2(\theta)\)
6. Cancel out the terms: \(4\cos(\theta) = 3\sin(\theta)\)
7. Express \(\cos(\theta)\) in terms of \(\sin(\theta)\): \(\tan(\theta) = \frac{4}{3}\)
8. Find \(\sin(\theta)\) and \(\cos(\theta)\):
   • \(\sin(\theta) = \frac{4}{5}\)
   • \(\cos(\theta) = \frac{3}{5}\)
9. Substitute back to find the range: \(R = \frac{u^2 \times \frac{3}{5} \times \frac{4}{5}}{g} = \frac{12u^2}{25g}\)
10. Equate the given range expression to find \( n \): \(\frac{nu^2}{25g} = \frac{12u^2}{25g}\)
11. Solve for \( n \): \(n = 12 \times 2 = 24\)
12. Thus, the value of \( n \) is \( 24 \).

Therefore, the correct option is 24.

Answer 2

The horizontal range \( R \) of a projectile launched with initial velocity \( u \) at an angle \( \theta \) with the horizontal is given by: \[ R = \frac{u^2 \sin 2\theta}{g} \] The maximum height \( H_{max} \) attained by the projectile is given by: \[ H_{max} = \frac{u^2 \sin^2 \theta}{2g} \] We are given that the horizontal range is three times the maximum height: \[ R = 3 H_{max} \] \[ \frac{u^2 \sin 2\theta}{g} = 3 \left( \frac{u^2 \sin^2 \theta}{2g} \right) \] \[ \frac{u^2 (2 \sin \theta \cos \theta)}{g} = \frac{3 u^2 \sin^2 \theta}{2g} \] We can cancel \( \frac{u^2}{g} \) from both sides (assuming \( u \neq 0 \)): \[ 2 \sin \theta \cos \theta = \frac{3}{2} \sin^2 \theta \] Assuming \( \sin \theta \neq 0 \) (i.e., the projectile is launched at an angle other than 0 or 180 degrees), we can divide by \( \sin \theta \): \[ 2 \cos \theta = \frac{3}{2} \sin \theta \] \[ \frac{\sin \theta}{\cos \theta} = \tan \theta = \frac{2}{3/2} = \frac{4}{3} \] Now we need to find the horizontal range \( R \) in terms of \( u \) and \( g \). We know \( R = \frac{u^2 \sin 2\theta}{g} = \frac{u^2 (2 \sin \theta \cos \theta)}{g} \). If \( \tan \theta = \frac{4}{3} \), 
we can consider a right-angled triangle where the opposite side is 4 and the adjacent side is 3. 
The hypotenuse is \( \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \). 
So, \( \sin \theta = \frac{4}{5} \) and \( \cos \theta = \frac{3}{5} \). 
Substituting these values into the expression for \( R \): \[ R = \frac{u^2 (2 \times \frac{4}{5} \times \frac{3}{5})}{g} = \frac{u^2 (\frac{24}{25})}{g} = \frac{24 u^2}{25 g} \] We are given that the horizontal range is \( R = \frac{nu^2}{25g} \). 
Comparing this with our result, we find that \( n = 24 \).