A child throws a ball with an initial speed of 8.0 m/s at an angle of 40° above the horizontal.The ball leaves her hand 1.0 m above the ground.At what angle below the horizontal does the ball approach the ground assuming no effects of air?

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Quizplus · Accepted Answer

The vertical component of the initial velocity is:$$v_{y_i} = v_i \sin 	heta = (8.0 	ext{ m/s}) \sin 40° = 5.20 	ext{ m/s}$$The horizontal component of the initial velocity is:$$v_{x_i} = v_i \cos 	heta = (8.0 	ext{ m/s}) \cos 40° = 6.13 	ext{ m/s}$$The time it takes for the ball to reach the ground is:$$t = \frac{2v_{y_i}}{g} = \frac{2(5.20 	ext{ m/s})}{9.81 	ext{ m/s}^2} = 1.05 	ext{ s}$$The horizontal distance traveled by the ball is:$$x = v_{x_i}t = (6.13 	ext{ m/s})(1.05 	ext{ s}) = 6.44 	ext{ m}$$The angle below the horizontal at which the ball approaches the ground is:$$	heta = 	an^{-1} \left(\frac{v_{y_f}}{v_{x_f}}ight) = 	an^{-1} \left(\frac{-v_{y_i}}{v_{x_i}}ight) = 	an^{-1} \left(\frac{-5.20 	ext{ m/s}}{6.13 	ext{ m/s}}ight) = -42°$$The negative sign indicates that the ball is moving downward. Therefore, the angle below the horizontal at which the ball approaches the ground is 42°.

A Child Throws a Ball with an Initial Speed of 8.0