- #1
Leo34005
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Homework Statement
Two students are on a balcony 23.4 m above the street. One student throws a ball, b1, vertically downward at 15.5 m/s. At the same instant, the other student throws a ball, b2, vertically upward at the same speed. The second ball just misses the balcony on the way down.
(a) What is the difference in time the balls spend in the air?
(b) What is the velocity of each ball as it strikes the ground?
velocity for b1
velocity for b2
(c) How far apart are the balls 0.480 s after they are thrown?
Homework Equations
s = ut + 1/2at
Not to sure
The Attempt at a Solution
I am considering g = 10 m / sec^2.
Motion of the first ball:
=>23.4 = (15.5)t + 10(t^2)/2
=>23.4 = 15.5 t + 5(t^2)
=>(t^2) + 3.1t - 4.68 = 0
=> t = (-3.1 + 5.3) / 2 = 0.6 sec
Motion of the second ball
=>23.4 = - (15.5)t + 10(t^2)/2
=>23.4 = -(15.5 t) + 5(t^2)
=>(t^2) - 3.1t - 4.68 = 0
=> t = (3.1 + 5.3) / 2 = 4.2 sec
Difference in time = 3.6 sec