InterviewSolution
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InterviewSolution
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A ball is thrown from a field with a speed of 12.0 m/s at an angle of `45^0` with the horizontal. At what distance will it hit the field again ? Take `g=10.0 m/s^2` |
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Answer» The horizontal range `= (u^2sin 2 theta)/g` `=((12 m/s)^2xxsin(2xx45^0))` `=(144 m^2/s^2)/(10.0m/s^2) = 14.4 m. Thus, the ball hits the field at 14.4 m from the point of projection. |
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