BITSAT 2015 Paper

Given : Initial velocity $\mu_0$ = 20 $\sqrt{2}$ m/s ; angle of projection θ = 45° Therefore horizontal and vertical components of initial velocity are $u_x$ = $20\sqrt{2}$ cos 450 = 20 m/s and $u_y$ = $20\sqrt{2}$ sin 45° = 20 m/s After 1s, horizontal component remains unchanged while the vertical component becomes $v_y$ = $u_y$ - gt Due to explosion, one part comes to rest. Hence, from the conservation of linear momentum, vertical component of second part will become $v'_y$ = 20 m/s Therefore, maximum height attained by the second part will be H = $h_1+h_2$ Using, h = ut + $\frac{1}{2} a t^2$ ⇒ $h_1$ = (20 ×1) - $\frac{1}{2} \times 10 \times (1)^2$ = 15 m a = g = 10 m/$s^2$ $h_2$ = $\frac{v_{y}^{\prime 2}}{2g}$ = $\frac{(20)^2}{2 \times 10}$ = 20 m H = 20 + 15 = 35 m