A stone is thrown horizontally outward from the top of a bridge. The stone is released 19.6 m above the street below. The initial velocity of the stone is 5 m/sec. Then the magnitude of velocity when it hits the floor of street is: (g=9.8 m/s2)

given height from where stone is projected=h=19.6m; acceleration due to gravity =9.8m/s2  initial velocity of stone=u=ux=5m/s;  magnitude of final velocity=v=? v=vx2+vy2 we know that, ux=vx=5 m/s here vertical component of initial velocity=uy=0 Vy2=uy2+2gh vy=2gh=2×9.8×19.6=19.6m/s v=vx2+vy2=52+19.62=20.22m/s