Two students are on a balcony 21.0 m above the street. One student throws a ball

Question

Two students are on a balcony 21.0 m above the street. One student throws a ball (ball 1) vertically downward at 10.6 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?
s

(b) What is the velocity of each ball as it strikes the ground?

(c) How far apart are the balls 0.900 s after they are thrown?
m

ball 1 magnitude m/s direction ---Select--- upward downward ball 2 magnitude m/s direction ---Select--- upward downward

Explanation / Answer

1st student> h=21, u = 10.6, time t1 to reach ground
h = ut1 + 0.5 gt1^2
21 = 10.6(t1) + 4.9^t1^2
gives t1 = 1.25[rejecting negative time root]
v1(ground) = u + gt1 = 10.6+9.8*1.25 = 22.86 m/s
--------------------------------------...
2nd student> h=21, u= - 10.6, time t2 to reach ground
h = ut2 + 0.5 gt2^2
21 = -10.6t2 + 4.9^t2^2
gives t2 = 3.417[rejecting negative time root]
v2(ground) = u + gt2 = -10.6 + 9.8*3.417 = 22.92 m/s
-----------------------------
a) time difference = 3.417 - 1.25 = 2.167 s
b) they reach with same velocities (nearly)
--------------------------
h1(0.9) = 10.6*0.9 + 4.9^0.9^2 = 13.509 m (down)

h2(0.9) = -10.6*0.9 + 4.9^0.9^2 = - 5.571 m (up)

h1(0.9) - h2(0.9) = 19.08
c) 19.08 m apart

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