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 \(\frac{nu^2}{25 g}\) , where value of n is : 

(Given \('g'\) is the acceleration due to gravity).

(1) 6

(2) 18

(3) 12

(4) 24

Answer 1

Correct option is : (4) 24

R = 3H

\(\frac{u^2 \ \sin2\theta}{g} = \frac{3u^2 \sin^2\theta}{2g}\)

\(2 \ sin\theta \cos\theta = \frac{3}{2} \sin \theta \sin\theta\)

\(\frac{4}{3} = \tan \theta\)

\(\theta = 53 ^\circ\)

\(R = \frac{u^2}{g}\times 2 \times \frac{4}{5} \times\frac{3}{5} = \frac{24}{25} \frac{u^2}{g}\)