What is the angle of projection to have a maximum range in kitti pull?if one strikes kitti pull with the speed of 98 ms-1 what is the maximum range achieved
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The angle of projection is 45° to have a maximum range .
Projected velocity=strikes velocity
maximum range = 2u²sin2©/g
=2×98×98×sin2×45°/9.8
=2×98×98×1/9.8
=2×98×10=1960m
Projected velocity=strikes velocity
maximum range = 2u²sin2©/g
=2×98×98×sin2×45°/9.8
=2×98×98×1/9.8
=2×98×10=1960m
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