A ball is thrown upward at an angle with a speed and direction such that it reaches a maximum height of 19.5 m above the point it was releas

Question

A ball is thrown upward at an angle with a speed and direction such that it reaches a maximum height of 19.5 m above the point it was released, with no appreciable air resistance. At its maximum height it has a speed of 18.5 m/s. With what speed was the ball released?a. 26.9 m/sb. 32.3 m/sc. 37.7 m/sd. 16.2 m/se. 21.5 m/s

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Euphemia 5 years 2021-07-18T10:01:51+00:00 2 Answers 220 views 0

Answers ( )

  1. Vodka
    0
    2021-07-18T10:03:00+00:00
    Reply

    Answer:

    a. 26.9 m/s

    Explanation:

    Parameters given:

    Height reached by ball, s = 19.5 m

    Speed at maximum height (final velocity), v = 18.5 m/s

    To find the initial speed of the ball, we use one of Newton’s equations of motion:

    v^2 = u^2 - 2gs

    where g = acceleration due to gravity

    The negative sign is present because the ball is thrown upward, hence, it moves against the gravitational force.

    Therefore:

    18.5^2 = u^2 - (2*9.8*19.5)\\\\\\u^2 = 18.5^2 + 382.2 \\\\\\u^2 = 724.4\\\\\\u = \sqrt{724.4} = 26.9 m/s

    The ball was released at a speed of 26.9 m/s.

  2. linhdan
    0
    2021-07-18T10:03:16+00:00
    Reply

    Answer:

    26.91 m/s

    Explanation:

    at maximum height it has a speed of 18.5 m/s which must be its horizontal component of the speed with which it was projected

    Vₓ = 18.5 m/s

    Vy = √ 2gh = √ (2 × 9.8 × 19.5 ) = 19.55 m/s

    V ( velocity with which it was projected ) =√ ( 18.5² + 19.55²) = 26.91 m/s

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