Mechanics

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Can you solve Q.1,thank you.

1 Answer

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  • 天同
    Lv 7
    5 years ago
    Favorite Answer

    (a)(i) Acceleration from A to B = 20/4 m/s^2 = 5 m/s^2

    Deceleration from C to D = 20/5 m/s^2 = 4 m/s^2

    (ii) Total distance travelled = area of trapezium = (6 + 15) x 20/2 m = 210 m

    (iii) Average speed = 210/15 m/s = 14 m/s

    (b)(i) Use equation v = u + at

    0 = 25 + (-10)t [take g = -10 m/s^2]

    i.e. t = 2.5 s

    (ii) Height from the starting point that the ball reached = 25^2/2g = 31.25 m

    Hence, max height from ground reached by the ball = (31.25 + 50) m = 81.25 m

    (iii) Height from starting point that the ball should reach

    = (95 - 50) m = 45 m

    Use equation: v^2 = u^2 + 2as

    0 = u^2 + 2(-10).(45)

    i.e. u = 30 m/s

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