Two students are on a balcony 20.0 mabove the street. One student throws a ball

Question

Two students are on a balcony 20.0 mabove the street. One student throws a ball (ball 1) verticallydownward at 11.1 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 2 m/s ball 2 3 m/s
(c) How far apart are the balls 0.800s after they are thrown?
4 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 2 m/s ball 2 3 m/s
(c) How far apart are the balls 0.800s after they are thrown?
4 m ball 1 2 m/s ball 2 3 m/s

Explanation / Answer

the differnce in time=time taken by the ball[ which is thrownupward] to reach the same point of projection =2*11.1/9.8=2.2653sec ball1 v*v=11.1*11.1+2*9.8*20=123.21+392=515.21m/sec v=22.69 m/sec ball2 ---------- v=-22.69m/sec in the time 0.800 sec the ball thrown up in thevertical direction only hence distance travelled by it in the upwarddirection=11.1*0.8-1/2*9.8*0.8*0.8=5.744m the distance travelled by the body that was moving dowanwards s=11.1*0.8+1/2*9.8*0.8*0.8=12.016m so they are 17.76 m apart

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