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15 June, 22:17

A projectile is launched horizontally from a height of 8.0 m. The projectile travels 6.5 m before hitting the ground.

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  1. 15 June, 22:44
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    You can find

    1) time to hit the ground

    2) initial velocity

    3) speed when it hits the ground

    Equations

    Vx = Vxo

    x = Vx * t

    Vy = Vyo + gt

    Vyo = 0

    Vy = gt

    y = yo - Vyo - gt^2 / 2

    => yo - y = gt^2 / 2

    1) time to hit the ground

    => 8.0 = g t^2 / 2 = > t^2 = 8.0m * 2 / 9.81 m/s^2 = 1.631 s^2

    => t = √1.631 s^2 = 1.28 s

    2) initial velocity

    Vxo = x / t = 6.5m / 1.28s = 5.08 m/s

    3) speed when it hits the ground

    Vy = g*t = 9.81 m/s * 1.28s = 12.56 m/s

    V^2 = Vy^2 + Vx^2 = (12.56 m/s) ^2 + (5.08 m/s) ^2 = 183.56 m^2 / s^2

    => V = √ (183.56 m^2 / s^2) = 13.55 m/s
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