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2 years ago

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 = 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

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