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4 June, 02:09

A dynamite blast at a quarry launches a rock straight upward, and 2.4 s later it is rising at a rate of 10 m/s. assuming air resistance has no effect on the rock, calculate its speed (a) at launch and (b) 5.1 s after launch.

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  1. 4 June, 02:33
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    Let u = the vertical speed at launch, m/s

    g = 9.8 m/s², acceleration due to gravity.

    Part (a)

    After 2.4 s, the vertical speed is 10 m/s.

    Therefore

    (10 m/s) = (v ft/s) - (9.8 m/s²) * (2.4 s)

    v = 10 + 9.8*2.4 = 33.52 m/s

    Answer: 33.52 m/s upward

    Part (b)

    After 5.1 s, the vertical speed is

    (33.52 m/s) - (9.8 m/s²) * (5.1 s) = - 16.46 m/s.

    This means that the rock reached the maximum height and began falling downward.

    Answer: 16.46 m/s downward
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