Phil leans over the edge of a cliff and throws a rock upward at 5 m/s. Neglecting air resistance, two seconds later the rock's speed is?

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  • MyRank
    Lv 6
    6 days ago

    Given,

    Time (t) = 2secInitial speed (u) = 5m/secAcceleration due to gravity (g) = 9.81m/sec²Final speed (v) = ?Using kinematic relation:-v = u – gtv = 5m/sec + (-g)t [∵ upward direction]= 5m/sec – 9.81m/sec² x 2sec= 5m/sec – 19.62m/sec= -14.62m/sec∴ final speed (v) = -14.62m/sec.

  • oubaas
    Lv 7
    2 weeks ago

    V = Vo-g*t = 5-9.8*2 = -14.6 m/sec

  • 2 weeks ago

    The rock has a vertical acceleration of -9.8 m/s^2. Use the following equation to solve this problem,

    vf = vi + a * t

    vf = 5 + -9.8 * 2 = -14.6 m/s

    The fact the final velocity is negative means the ball is moving downward. I hope this is helpful for you.

  • Ash
    Lv 7
    2 weeks ago

    v = u + at

    lets consider upward velocity as positive and downward as negative, we get

    v = 5 + (-9.8)(2)v = -14.6 m/sThe velocity after 2 sec is 14.6 m/s downwards

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  • Anonymous
    2 weeks ago

    5 m/sec - 2 sec * 9.82 m/sec^2

    Can you handle it from here?

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