Question

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

Answer 1

Answer:

122.17 m/s

Explanation:

x cos 40 = horizontal velocity

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

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 x

Equate the two times from above to solve for x

1958.11/ x  =  .13118 x

x = 122.17 m/s