A dynamite blast at a quarry launches a rock straight upward, and 1.7 s later it is rising at a rate of 17 m/s. Assuming air resistance has

Question

A dynamite blast at a quarry launches a rock straight upward, and 1.7 s later it is rising at a rate of 17 m/s. Assuming air resistance has no effect on the rock, calculate its speed (a) at launch and (b) 4.9 s after launch.

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Dâu 5 years 2021-08-28T14:01:52+00:00 1 Answers 32 views 0

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  1. vankhanh
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    2021-08-28T14:02:53+00:00
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    Answer:

    (a) v_0= 34m/s

    (b) v=14m/s

    Explanation:

    We can derive the initial speed of the rock from the equation of the speed in function of the time:

    v=v_0-gt\\\\\implies v_0=v+gt

    Using the given values for the speed at time t=1.7s, we get:

    v_0=17m/s+(9.8m/s^{2})(1.7s)=34m/s

    In words, the speed of the rock at launch is 34m/s (a).

    Next, we use this to calculate the speed at t=4.9s:

    v=v_0-gt\\\\v= 34m/s-(9.8m/s^{2})(4.9s)=-14m/s

    This means that the speed of the rock at 4.9s after the launch is 14m/s (b), and the negative sign means that it is moving downwards.

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