A ball is thrown upward from a cliff. The ball rises to its maximum height and then falls into the valley below. Suppose the ball is thrown

Question

A ball is thrown upward from a cliff. The ball rises to its maximum height and then falls into the valley below. Suppose the ball is thrown upward at 30m/s and lands 9s after it is thrown. What is the height of the cliff?

in progress 0
Adela 3 months 2021-08-01T13:48:27+00:00 1 Answers 1 views 0

Answers ( )

    0
    2021-08-01T13:50:18+00:00

    Answer:

    The height of the cliff is approximately 127.31 meters

    Explanation:

    The given parameters are;

    The upward speed of the ball = 30 m/s

    The time after the ball is thrown before it lands = 9 s

    We have;

    The time to maximum height;

    v = u – g·t

    Where;

    v = The velocity at maximum height = 0 m/s

    u = Initial velocity = 30 m/s

    g = The acceleration due to gravity = 9.81 m/s²

    ∴ 0 = 30 – 9.81 × t

    t = 30/9.81 = 3.06 s

    The maximum height can be obtained from;

    v² = u² – 2·g·s

    v = 0 m/s

    u² = 2·g·s

    30² = 2 × 9.81 × s

    s = 30²/(2×9.81) = 45.87 m

    The time from maximum height to landing = 9 – 3.06 = 5.942 s

    The height to landing

    s = u·t + 1/2·g·t².

    Here, u = 0 m/s

    s = 0×5.942 + 1/2×9.81×5.942² = 173.1766 m

    s = 173.1766 m.

    The height of the cliff = 173.1766 m. – 45.87 m. ≈ 127.31 m.

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )