Aball is thrown horizontally from the top of a building 0.10 km high. the ball strikes the ground at a point 65 m horizontally away from and below the point of release. what is the speed of the ball just before it strikes the ground? (a) 43 m/s (b) 47 m/s (c) 39 m/s (d) 36 m/s (e) 14 m/s

47 ms

well the time it takes to fall 100m is the same time it takes to travel 65m horizontally.

The time to fall vertically, t is sqrt(2d/g). [this comes from d = 1/2at^2]

so t = sqrt(2*100/9.8) = 4.52s.

The vertical speed at that time is g*t = 9.8*4.52 = 44.3m/s

The horizontal speed is the horizontal distance over the same 4.52s, = 65/4.52 = 14.4m/s.

so the final velocity is = sqrt(44.3^2 + 14.4^2) = 47.ms

f=gm1m2/d^2 it d

f is force g is gravitaiton constant, m1 is mass of first object, m2 is mass of second object. d is distance between two object