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February 28, 2015

February 28, 2015

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

(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 10:57amThrown 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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