Two students are on a balcony 17.6 m above the street. One student throws a ball

Question

Two students are on a balcony 17.6 m above the street. One student throws a ball vertically downward at 15.7 m/s. At the same instant, the other student throws a ball vertically upward at the same speed. The second ball just misses the balcony on the way down. (Assume the positive direction is upward.)
(a) What is the difference in the time the balls spend in the air?
_______ s

(b) What is the velocity of each ball as it strikes the ground?
ball thrown downward
__________ m/s
ball thrown upward
__________m/s

(c) How far apart are the balls 0.800 s after they are thrown?
_____________m

Explanation / Answer

d = v0*t + (1/2)at^2 d = (15.7)t - 4.9t^2 -17.6 = 15.7t - 4.9t^2 (1) -17.6 = - 15.7t - 4.9t^2 (2) (1) t = 4.08 (2) t = 0.88 a) Difference = 4.08 - 0.88 = 3.2 seconds b) Use the formula: vf = v0 - gt The initial velocity is zero leaving -gt. For (1) - (9.8)*(4.08) = - 39.98 m/s For (2) - (9.8)*(0.88) = - 8.62 m/s c) Plug in 0.800 seconds in each equation then consider its difference

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