Answer : 0.50
\(X = \frac{3R}{2}\) as Xcm = R
\(R = \frac{{u_2 \sin(2\theta)}}{g}\)
\(= \frac{50}{10} = 5\)
\(βX = \frac{3R}{2} = \frac{15}{2} = 7.5 m\)
Answer: 7.50
A particle is projected at an angle of \( 30^\circ \) from horizontal at a speed of 60 m/s. The height traversed by the particle in the first second is \( h_0 \) and height traversed in the last second, before it reaches the maximum height, is \( h_1 \). The ratio \( \frac{h_0}{h_1} \) is __________. [Take \( g = 10 \, \text{m/s}^2 \)]