Question

Two physics students are on a balcony 19.6 m above the street. One student throws a ball vertically downward at 14.7 m/s; at the same instant, the other student throws a ball vertically upward at the same speed. The second balljust misses the balcony on the way down. (a) What is the difference in the two balls' time in the air? (b)What is the velocity of each ball as it strikes the ground? (c) How far apart are the balls 0.8 s after they are thrown?

Answer 1

A ) The first ball:
h = v o * t - g t² / 2
19.6 = 14.7 t + 9.8 t²/2
4.9 t² + 14.7 t - 19.6 = 0
t 1/2 = (-14.7 +/- √216.09+384.16 ) / 9.8 = 9.8/9.8 = 1 s
The second ball:
v = v o - g t
0 = 14.7 - 9.8 t
9.8 t = 14.7
t = 1.5 ( upward )
h = 19.6 + 14.7 - 9.8 * 1.5 = 30.625 m 
30.625 = 9.8 * t²/2
t² = 6.25
t = 2.5 s ( downward )
1.5 s + 2.5 s = 4 s
The difference: 4 s - 1 s = 3 s.
B )  The first ball:
v = v o + g t = 14.7 + 9.8 * 1 = 24.5 m/s
The second ball:
v = g * t = 9.8 * 3 = 24.5 m/s
C )  d 1 = 14.7 * 0.8 + 9.8 * 0.8²/2 = 11.76 + 3.136 = 14.896 m
d 2 = 14.7 * 0.8 - 9.8* 0.8²/2 = 11.76 - 3.136 = 8.624 m 
d 1 + d 2 = 14.896 + 8.624 = 23.52 m