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 17.7 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?
1 s

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

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

Explanation / Answer

a. For the ball thown downwards u = -17.7 m/s a = -9.81 m/s^2 s = -17.6m t =? s = ut + 0.5at^2 -17.6 = -17.7t + 0.5(-9.81)t^2 4.905t^2 + 17.7t - 17.6 = 0 t = 0.81175 or -4.4203 (NA) time ball spends in the air = 0.81175s For the ball thrown upwards u = 17.7 m/s a = -9.81 m/s^2 s = -17.6m t =? s = ut + 0.5at^2 -17.6 = 17.7t + 0.5(-9.81)t^2 4.905t^2 - 17.7t - 17.6 = 0 t = -0.81175(NA) or 4.4203 time ball spends in the air = 4.4203s Time difference = 4.4203 - 0.81175 = 3.61s b. v = u + at for ball thrown downward: v = 17.7 + 9.8*0.81175 = 25.655 m/s for ball thrown upward: v = 25.655 m/s c. For the ball thrown downward u = -17.7 m/s a = -9.81 m/s^2 t = 0.80s s = ? s = ut + 0.5at^2 Dispacement = -17.7(0.8) + 0.5(-9.81)(0.8^2) = -17.299 = 17.299m downward For the ball thrown upward u = 17.7 m/s a = -9.81 m/s^2 t = 0.80s s = ? s = ut + 0.5at^2 Dispacement = 17.7(0.8) + 0.5(-9.81)(0.8^2) = 11.021 = 11.021m upward Distance apart = 11.021 + 17.299 = 28.3m

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