A 35 g steel ball is held by a ceiling-mounted electromagnet 3.8 m above the floor. A compressed-air cannon sits on the floor, 4.8 m to one

Question

A 35 g steel ball is held by a ceiling-mounted electromagnet 3.8 m above the floor. A compressed-air cannon sits on the floor, 4.8 m to one side of the point directly under the ball. When a button is pressed, the ball drops and, simultaneously, the cannon fires a 25 g plastic ball. The two balls collide 1.4 m above the floor. What was the launch speed of the plastic ball?

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Khải Quang 4 years 2021-08-08T17:35:47+00:00 1 Answers 412 views 1

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    2021-08-08T17:36:57+00:00

    Answer:

    Explanation:

    Let v is the launch speed of the plastic ball and the angle of projection is θ.

    So, in horizontal direction

    v Cosθ x t = 4.8 …. (1)

    In th evertical direction

    1.4 = v Sin θ x t – 0.5 gt²     …. (2)

    As , v Sin θ x t = 3.8      …. (3) , put in equation (2)

    1.4 = 3.8 – 4.9 t²

    t = 0.7 s

    Put in (1) and (3)

    v Cosθ x 0.7  = 4.8

    v Cosθ = 6.86

    and v Sinθ x 0.7 = 3.8

    v Sinθ = 5.43

    Now

    v^{2}\left ( Sin^{2}\theta +Cos^{2}\theta  \right )=5.43^{2}+6.86^{2}

    v = 8.75 m/s

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