Question #101105
A cricketer throws a 147g ball by exerting a force of 7.5N for 0.84s. If the launch angle is 39° from horizontal, calculate the range of the ball.
1
Expert's answer
2020-01-08T09:48:39-0500

Find initial velocity


v0=Fmt0v_0=\frac{F}{m}t_0

It's known problem - throwing the body at an angle to the horizon


x=v0cos(α)tx=v_0cos(\alpha)ty=v0sin(α)tgt22y=v_0sin(\alpha)t-\frac{gt^2}{2}

From this equation


x=2v02sin(α)cos(α)gx=\frac{2v_0^2sin(\alpha)cos(\alpha)}{g}

Calculating


x=2(7.50.840.147)2sin(39)cos(39)10=183(m)x=\frac{2\cdot (\frac{7.5\cdot 0.84}{0.147})^2 \cdot sin(39)\cdot cos(39)}{10}=183 (m)


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