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Homework Help: physics

Posted by morgan on Tuesday, September 11, 2012 at 12:14am.

Two students are on a balcony 24.4 m above the street. One student throws a ball, b1, vertically downward at 20.3 m/s. At the same instant, the other student throws a ball, b2, vertically upward at the same speed. The second ball just misses the balcony on the way down.
(a) What is the difference in time the balls spend in the air?

(b) What is the velocity of each ball as it strikes the ground?
velocity for b1

velocity for b2

(c) How far apart are the balls 0.510 s after they are thrown?

• physics - Henry, Wednesday, September 12, 2012 at 8:52pm

  The First Ball:
  d = Vo*t + 0.5g*t^2 = 24.4 m.
  20.3t + 4.9t^2 = 24.4
  4.9t^2 + 20.3t - 24.4 = 0
  Use Quadratic Formula:
  Tf = 0.973 s.=Fall time or time in air.
  
  The 2nd Ball:
  Tr = (V-Vo)/g = (0-20.3)/-9.8 = 2.07 s.=
  Rise time.
  h = ho + Vo*t + 0.5g*t^2
  h=24.4 + 20.3*2.07 - 4.9(2.07)^2=45.4 m
  Above gnd.
  d = Vo*t + 0.5g*t^2 = 45.4 m.
  0 + 4.9t^2 = 45.4
  t^2 = 9.27
  Tf = 3.04 s.
  Tr + Tf = 2.07 + 3.04 = 5.1 s. = Time
  in air.
  
  a. 5.1 - 0.973 = 4.14 s.
  
  b. V1^2 = Vo^2 + 2g*d.
  V1^2 = (20.3)^2 + 19.6*24.4 = 890.3
  V1 = 29.8 m/s.
  
  V2^2 = 0 + 19.6*45.4 = 889.8
  V2 = 29.8 m/s.
  
  c. h1 = ho - (Vo*t + o.5g*t^2.)
  h1 = 24.4 - (20.3*0.51 + 4.9(0.51)^2
  h1 = 24.4 - 11.63 = 12.8 m. Above gnd.
  
  h2 = ho + (20.3*0.51 - 4.9(0.510^2
  h2 = 24.4 + 9.08 = 33.5 m. Above gnd.
  
  Da = h2 - h1 = 33.5 - 12.8 = 20.7 m.
  Apart.
  

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