Physics

A ball is thrown straight upward and returns to the thrower's hand after 2.30 s in the air. A second ball is thrown at an angle of 42.0° with the horizontal. At what speed must the second ball be thrown so that it reaches the same height as the one thrown vertically?

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asked by Harley
  1. first ball stops in 2.3/2 = 1.15 seconds
    0 = Vi - g t
    Vi = 9.81 (1.15) = 11.3 m/s

    second ball has same Vi
    Vi = V sin 42
    V = 11.3/sin 32 = 21.3 m/s

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    posted by Damon
  2. vf=vi+at, 0=vi+)(9,81)(1,15)

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    posted by peace
  3. damon how you go from sin 42 to sin 32

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    posted by none

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