Posted by **fsfsf** on Wednesday, March 2, 2011 at 11:21pm.

two students are on a balcony 20.6 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 ball just misses the balcony on the way down. (Assume the positive direction is upward.)

(a) What is the difference in the time the balls spend in the air?

s

(b) What is the velocity of each ball as it strikes the ground?

ball thrown downward

m/s

ball thrown upward

m/s

(c) How far apart are the balls 0.800 s after they are thrown?

- physics -
**tchrwill**, Thursday, March 3, 2011 at 10:57am
Thrown down:

From Vf^2 = Vo^2 + 2gh, Vf = 24.89m/s.

From Vf = Vo + gt, t = 1.04sec.

Thrown upward:

From Vf = Vo - gt, t = 1.5sec. to Vf = 0

From h = Vo(t) - gt^2/2, h = 11.025m.

From peak height to ground = 31.625m.

From Vf^2 = Vo^2 + 2gh, Vf = 24.896m/s.

Distance traveled by ball thrown upward after .8 sec. derives from d = Vo(t) - gt^2/2 or d = +8.624m.

Distance traveled by ball thrown downward after .8 sec. derives from d = Vo(t) + gt^2/2 or d = 14.89m.

a) Difference in air time is 4.04 - 1.04 = 3sec.

b) Vf thrown downward = 24.89m/s

...Vf thrown upward = 24.89m/s

c) Therefore, after .8 sec., the balls are 23.52m apart.

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