You drop a ball from a window on an upper floor of a building and it is caught by a friend on the ground when the ball is moving with speed vf. You now repeat the drop, but you have a friend on the street below throw another ball upward at speed vf exactly at the same time you drop your ball from the window. The two balls are initially separated by 28.7 m. (a) At what time do they pass each other? (b) At what location do they pass each other relative the window?

Question

Isabelle

Physics

27 April, 10:55

You drop a ball from a window on an upper floor of a building and it is caught by a friend on the ground when the ball is moving with speed vf. You now repeat the drop, but you have a friend on the street below throw another ball upward at speed vf exactly at the same time you drop your ball from the window. The two balls are initially separated by 28.7 m. (a) At what time do they pass each other? (b) At what location do they pass each other relative the window?

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Keaton Arias · Answer 1

Let they will meet at y above surface ground.

Vf² = v ² + 2 g H

Vf² = 0 ² + 2 (9.8) (28.7)

Vf = 23.7175 m/s

analyzing for the first ball,

y = Yo + ½ g t²

y = 28.7 + ½ (-9.8) t²

and for the second one,

y = Vf t + ½ g t²

y = 23.7175 t + ½ (-9.8) t²

and then. they meet each other,

y = y

28.7 + ½ (-9.8) t² = 23.7175 t + ½ (-9.8) t²

t = 1.21008 sec

y = 23.7175 t + ½ (-9.8) t²

y = 23.7175 (1.21008) + ½ (-9.8) (1.21008) ²

y = 21.5250 m