Two students are on a balcony 18.4 m abovethe street. One student throws a ball
ID: 1662550 • Letter: T
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
given the velocity of the balls are11.9m/s; height of the balcony 18.4m. here is the main point of this question that when ever ny bodythrown upward with velocity v then in return time it strick theground with the same velocity v(u can see it easily applying theformula of kinemetics). so the velocity of the 2nd ball when it return to the balcony issame as 11.9m/s. ?T = t2+t1; T=(time taken by the ball 2 to travel from balcony to theground); t2=(time taken by ball 2 from balcony to the top and again tobalcony). t1=(time taken by ball 2 and also ball 1 from balcony to the theground). a) the difference in the two ball's time in the air=T-t1=t2; but t2 = 2usin90/g = 2*11.9*1/9.81 = 2.426s. b) as the velocity of the ball at the balcony is same there velocityat the ground is also same; v2-11.92=2*9.81*18.4; v=22.41m/s. c) after 0.8s ball 1 is : S1=11.9*0.8+1/2*9.81*0.8*0.8=12.6592m. for ball 2 : S2=11.9*.8-1/2*9.81*.8*.8=6.3808m; so the distance between balls after 0.8s is12.6592+6.3808=19.04m (because ball 1 is going down and ball 2 is moving up).
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.