Two people are standing on the edge of a building that is 42 meters high. One person throws a tennis ball straight downward at a speed of 16 m/s. At the same exact time, the other person throws a tennis ball straight upward at 16 m/s. How long after the first tennis ball lands will the second tennis ball arrive at the ground?

Answer : 1.67 sec

Explanation:

Time calculation for person 1

Height of the building = 40meters

Initial velocity = 16m/s

Final velocity = 0m/s

Accelaeration = 9.8m/s2

Using first equation of motion

V=u+at

0 = 16 + 9.8t

-16/-9.8 = t

Time = 1.63 seconds

Time calculation for person 2

While the ball is going up

Height of the building = 40meters

Initial velocity = 16m/s

Final velocity = 0m/s

Accelaeration = - 9.8m/s2

Using second equation of motion

v2 - u2 = 2as

0-256 = - 19.6s

S = - 256/-19.6 = 13.06m

Total distance. = 13.06 + 40 = 53.06 meters

Using third equation of motion

S = ut + ½ * a * t*t

S = 0 + ½ * 9.8*t*t

T2 = 53.06/4.9

T2 = 10.83

T =  10.83 = 3.3 sec

Diffenrece of time = 3.3 - 1.63 = 1.67 sec