Question

A juggler throws two balls in the air. She throws the first ball up with a velocity of 9. 8 m/s. She throws the second ball up with a velocity of 14. 7 m/s. How long after catching the first ball will she catch the second?.

Answer 1

t = (v2 - v1) / a

Since the time to fall equals the time to rise

t1 = 2 * 9.8 / 9.8 s = 2 sec

t2 = 2 * 14.7 / 9.8 = 3 sec

The second ball will be caught 1 sec after the first ball if they both start upwards at the same time.