Given : Initial velocity µ0 = 20 √2 m/s ; angle of projection θ = 45° Therefore horizontal and vertical components of initial velocity are ux = 20√2 cos 450 = 20 m/s and uy = 20√2 sin 45° = 20 m/s After 1s, horizontal component remains unchanged while the vertical component becomes vy = uy - 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 = h1+h2 Using, h = ut +