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Question

A ball is thrown from ground level of a field with a speed of 12.0 m/s at an angle of 45 with the horizontal. At what distance will it hit the field again? Take g = 10.0 m/s2.


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Solution

The horizontal range = u2sin2θg

= (12m/s)2×sin(2×45)10m/s2

= 144m2/s210.0m/s2 = 14.4m

Thus, the ball hits the field at 14.4 m from the point of projection.


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