Two students are on a balcony 19.4 mabove the street. One student throws a ball

Question

Two students are on a balcony 19.4 mabove the street. One student throws a ball (ball 1) verticallydownward at 16.6 m/s; at the sameinstant, the other student throws a ball (ball 2) vertically upwardat the same speed. The second ball just misses the balcony on theway 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 m/s direction up or downward ball 2 magnitude m/s direction up or downward
(c) How far apart are the balls 0.800s after they are thrown?
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 m/s direction up or downward ball 2 magnitude m/s direction up or downward
(c) How far apart are the balls 0.800s after they are thrown?
m ball 1 magnitude m/s direction up or downward ball 2 magnitude m/s direction up or downward

Explanation / Answer

h = 19.4 m, v = 16.6 m/s, choose downward as positive direction for ball 1: time = t1 h = vt1 + gt12/2 for ball 2: time = t2 h = -vt2 + gt22/2 vt1 + gt12/2 =-vt2 + gt22/2 v(t1 + t2) = g(t22- t12)/2 v(t1 + t2) = g(t2 -t1)(t1 + t2)/2 v = g(t2 - t1)/2 t2 - t1 = 2v/g = 3.39 s b) use vf2 = vi2 +2gh vf = (vi2 + 2gh) for ball 1, vi = +16.6 m/s, for ball 2, vi =-16.6 m/s so both has same vf = 25.6 m/s c) t = 0.800 s y1 = vt + gt2/2 y2 = -vt + gt2/2 so y1 - y2 = 2vt = 26.56 m

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