Question

A ball is thrown from ground level of a field with a speed of 12.0 m/s at an angle of ${45}^{\circ }$ with the horizontal. At what distance will it hit the field again? Take g = 10.0 m/$s^2$.

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Answer 1

The horizontal range = $\frac{u^{2}sin2\theta }{g}$

= $\frac{(12m/s{)}^2\times sin(2\times {45}^{\circ })}{10m/s^2}$

= $\frac{144m^2/s^2}{10.0m/s^2}$ = 14.4m

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