Two students are on a balcony 18.4 m abovethe street. One student throws a ball

Question

Two students are on a balcony 18.4 m abovethe street. One student throws a ball (ball 1) vertically downwardat 11.9 m/s; at the same instant, theother student throws a ball (ball 2) vertically upward at the samespeed. The second ball just misses the balcony on the way down. (a) What is the difference in the two ball'stime in the air?
1 s

(b) What is the velocity of each ball as it strikes the ground? ball 1 magnitude 2 m/s
direction 3---Select---upwarddownward ball 2 magnitude 4 m/s
direction 5---Select---upwarddownward
(c) How far apart are the balls 0.800 safter they are thrown?
6 m (a) What is the difference in the two ball'stime in the air?
1 s

(b) What is the velocity of each ball as it strikes the ground? ball 1 magnitude 2 m/s
direction 3---Select---upwarddownward ball 2 magnitude 4 m/s
direction 5---Select---upwarddownward
(c) How far apart are the balls 0.800 safter they are thrown?
6 m ball 1 magnitude 2 m/s
direction 3---Select---upwarddownward ball 2 magnitude 4 m/s
direction 5---Select---upwarddownward

Explanation / Answer

h = 18.4 m, v = 11.9 m/s; choose downward as positive direction. a) for student 1: h = vt1 + gt12/2     (1) for student 2: h = -vt2 + gt22/2 (2) (2) - (1) 0 = -v(t2 - t1) + g(t22- t12)/2 v(t2 - t1) = g(t22 -t12)/2 2v = g(t2 - t1) t2 - t1 = 2v/g = 2.43 s b) v12 = v2 + 2gh v1 = (v2 + 2gh) = 22.4m/s   direction: dowaward v22 = (-v)2 + 2gh v1 = (v2 + 2gh) = 22.4m/s    direction: dowaward c) t = 0.800 s height of the 1st ball y1 = h - (vt +gt2/2) height of the 2nd ball y2 = h - [(-v)t +gt2/2] y2 - y1 = 2vt = 19.04 m

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