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3 September, 10:40

A ball is launched vertically upward from the edge of a cliff. The ball reaches its maximum height 1.6 seconds after launch. Barely missing the edge of the cliff as it falls downward, the ball strikes the ground 6 seconds after being launched. a) What was the ball's initial velocity? b) What is the maximum height the ball reached above the cliff? c) How tall is the cliff?

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  1. 3 September, 11:05
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    Answer: a) 60m/s b) 360m c) 180m

    Explanation:

    Projectile motion occurs when an object launched into space falls freely under the influence of gravity.

    If the time of flight (T) = 2Usin (theta) / g where;

    U is the velocity of the object

    theta is the angle at which the object was thrown

    g is the acceleration due to gravity

    Given T = 6 seconds

    theta = 90° (object launched upwards)

    g = 10m/s²

    U=?

    Substituting in the formula we have;

    6 = Usin90°/10

    60 = Usin90°

    since sin90° = 1

    U = 60m/s

    The balls initial velocity is 60m/s.

    b) Maximum Height reached (H) = U²sin²theta/g

    Given U = 60m/s, theta = 90° g = 10m/s²

    H = 60² (sin90°) ²/10

    H = 3600/10

    H = 360m.

    The maximum height reached is 360m

    c) at maximum height, v = 0

    Using v² = u² - 2gh

    0 = 60² - 2 (10) h

    -3600 = - 20h

    h = 3600/20

    h = 180m

    The cliff is 180m tall
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