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April 18, 2014

April 18, 2014

Posted by **B** on Monday, August 29, 2011 at 2:47am.

(a) What is the difference in the two ball's time in the air?

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

ball 1

ball 2

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

cannot seem to figure out how to find part a) i cannot get the second balls time.

- physics -
**Henry**, Wednesday, August 31, 2011 at 3:31pma. d1 = 12.2t + 0.5gt^2 = 18.3m.

12.2t + 0.5*9.8t^2 = 18.3,

4.9t^2 + 12.2t - 18.3 = 0,

Solve using Quadratic Formula and get:

t1 = 1.054, and -3.54s.

Use positive value of t:

T1 = 1.054s = Time in air.

Vf = Vo + gt,

t2(up) = (Vf-Vo) / g,

t2(up) = (0-12.2) / -9.8 = 1.24s.

Vf^2 = Vo^2 + 2gd,

d2(up) = (Vf^2-Vo^2) / 2g,

d2(up) = (0-148.8) / -19.6 = 7.59m.

D2 = 7.59 + 18.3 = 25.89m from ground.

d = Vo*t + 0.5gt^2 = 25.89,

0 + 0.5*9.8*t^2 = 25.89,

4.9t^2 = 25.89,

t^2 = 5.28,

t(down) = 2.3s.

T2 = 1.24 + 2.3 = 3.54s in air.

T2 - T1 = 3.54 - 1.054s difference.

b. Vf^2 = Vo^2 + 2gd,

V1^2 = 12.2 + 2*9.8*18.3 = 370.88,

V1 = 19.26m/s.

V2^2 = 0 + 2*9.8*25.89 = 507.44,

V2 = 22.53m/s.

c. d1 = Vo*t + 0.5gt^2

d1 = 12.2*0.8 + 0.5*9.8*(0.8)^2 = 12.90m below the balcony.

d2=12.2*0.8 + 0.5*(-9.8)(0.8)^2 = 6.62m above the balcony

D = 6.62 + 12.90 = 19.52m apart.

- physics -
**Mercy**, Monday, March 25, 2013 at 1:03amA particle (a) is projected horizontally at 50m/s from the top of a tower 100m high at same instant aoother particle is projected from bottom of the tower in same vertical plain at 100m/s with elevation 60^0.after how maoy seconds will the particle collide

- physics -
**Mercy**, Monday, March 25, 2013 at 1:03amA particle (a) is projected horizontally at 50m/s from the top of a tower 100m high at same instant aoother particle is projected from bottom of the tower in same vertical plain at 100m/s with elevation 60^0.after how maoy seconds will the particle collide

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