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Sever21 [200]
2 years ago
7

A shell is launched with an initial velocity at an angle of 40.0° above horizontal

Physics
1 answer:
olasank [31]2 years ago
7 0

Answer:

122.17 m/s

Explanation:

x cos 40 = horizontal velocity

1500 m / x cos 40   = time in the air =<u> 1958.11 / x</u>

x sin 40 = vertical velocity

 find when shell vertical velocity = 0 (this is max height....1/2 way through its flight)  , the time when it hits the ground will be twice this...  

   0 =  x sin 40 - 9.8 t

  t =  x sin40 / 9.8        time in the air is twice this = .13118<u> x</u>

<u />

Equate the two times from above to solve for x

1958.11/ x  =  .13118 x

x = 122.17 m/s

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