Two students are on a balcony 20.5 m above the street. One student throws a ball

Question

Two students are on a balcony 20.5 m above the street. One student throws a ball (ball 1) vertically downward at 11.1 m/s; at the same instant, the other student throws a ball (ball 2) vertically upward at the same speed. The second ball just misses the balcony on the way down.
(a) What is the difference in the two ball's time 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--- upward downward
ball 2 magnitude 4 m/s
direction 5 ---Select--- upward downward

(c) How far apart are the balls 0.600 s after they are thrown?
6 m

Explanation / Answer

Given: initial velocity of the two balls u = 11.1 m/sec vertically downwards : from the kinematic raltions v2 - u2 = 2as final velocity of ball 1 before reach the ground v = sqrt (u2+2as)                                                                              = sqrt [(11.1)2+2*(9.8 m/sec2 *20.5 m )]                                                                              = 22.91 from kinematic relations v = u+at time in air                        t = v-u /a                                            = (22.91-11.1) / (9.8 m/sec2)                                            = 1.205 sec vertically upwards: time of flight t = 2u /g                        = (2*11.1 m/sec) / 9.8 m/sec2                        = 2.26 sec (1) difference in their timing in air t = 2.26 sec - 1.21 sec                                                       = 1.05 sec                                                       = 1 sec (2) final velocity of ball 1 v = 22.91 m/sec   (from the above) final valocity of ball 2 : maximum height reached by the ball H = u2/2g                                                              = (11.1)2 / (2*9.8 m/sec2 )                                                              = 6.26 m total height H = 6.26 m + 20.5 m                       = 26.78 m v = u+at     = (11.1 m/sec) +*(9.8m/sec2)(2.26 sec )     = 33.24 m/sec .............................................................................................................. for ball 1 s = ut +1/2at2    = (11.1)(0.6) +1/2*(9.8m/sec2)(0.6)2    = 8.424 m for ball 2 s = ut-1/2gt2     = (11.1)(0.6) -1/2*(9.8m/sec2)(0.6)2    = 4.89 m distance between them d = 4.89+8.424 m                                          = 13.3 m     = (11.1)(0.6) -1/2*(9.8m/sec2)(0.6)2    = 4.89 m distance between them d = 4.89+8.424 m                                          = 13.3 m    = 4.89 m distance between them d = 4.89+8.424 m                                          = 13.3 m

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