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

increased  this is because he is going up and that is making it so that it is going against gravity so if the elevator was to fail the elevator will go down which turns that potential energy to cenetic energy