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    I'm going through OCR past exams, and on the June 07 exam, question 2, I managed to confuse myself slightly. The question is:

    "Calculate the range on a horizontal plane of a small stone projected from a point on the plane with speed 12 m/s at an angle of elevation of 27 degrees."

    So I got the answer pretty quickly, by finding how long it took to get to the top of its path where v=o, vertically. Using v^2 =u^2 + 2as, where v=0, u=12sin27, a=-9.8, I found s to be 1.514. From there I just found the time according to v=u+at, doubled it for time of flight, and plugged that time into the horizontal s=ut+1/2 at^2.

    My question is from the part where I found the vertical displacement when v=o. Surely there are two points when v=o; at the top of its path, and at the end of the path. Should I not get another answer for s (0)?
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    There's only one point where the vertical velocity is zero that you need to consider - at the top of its flight.

    When it comes down again, just before reaching the ground the vertical velocity will not be zero, and as it hits the ground and stops, the equations of motion you are using no longer apply - you are no longer dealing with the same constant acceleration.
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    Oh I see, that makes sense, thanks
 
 
 
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Updated: April 13, 2011
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