if a ping pong ball drops from the top of the building that is 168 feet high, how many seconds will it take for the ball to hit the ground?
16t^2 = 168
find t
Well, let's assume that it is a lead ping pong ball so we can ignore air drag, and since you said 168 feet and not meters I will use the ancient units and g is 32 ft/s^2
easy way:
potential energy at top = kinetic energy at bottom
m g h = (1/2) m v^2
32 * 168 = (1/2) v^2
v^2 = 10752
v = 104 ft/second at ground
since the acceleration is constant and v= 0 at the top, the average speed = 104/2 = 52 ft/s
time = distance / speed = 168/52 = 3.23 seconds
============
alternately
a = - g = -32 ft/s^2
v = -g t
h = 168 - 16 t^2
t^2 = 10.5
t = 3.24 seconds