a 9 kg bowling ball rolls off a table and strikes the ground. if the ball isi n the air for 0.5 seconds, how fast is the ball moving when it hits the ground?

Answer 1

One of the formulas for uniformly accelerated motion (UAM):
h = vo*t - (1/2)gt^2

Ball falls freely, so initial velocity (vo) is zero:
h = 0 - (1/2)(-9.8)(0.5^2)
h = 1.225 meters

Since we're looking for the terminal velocity (vf) we can use the formula,
vf^2 - vo^2 = 2gd
vf^2 - 0 = 2*9.8*1.225
vf^2 = 24.01
vf = 4.9 m/s