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28 July, 12:36

Suppose you throw a ball vertically upward with a speed of 49 m/s. Neglecting air friction, what would be the height of the ball be 2.0 second later?

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  1. 28 July, 12:43
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    78.4 m

    Explanation:

    Using newton's equation of motion,

    S = ut + 1/2gt² ... Equation 1

    Where S = Height, t = time, u = initial velocity, g = acceleration due to gravity.

    Note: Taking upward to be negative, and down ward positive

    Given: u = 49 m/s, t = 2.0 s, g = - 9.8 m/s²

    Substitute into equation 1

    S = 49 (2) - 1/2 (9.8) (2) ²

    S = 98 - 19.6

    S = 78.4 m

    Hence the height of the ball two seconds later = 78.4 m
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