How long do the balls take to hit the ground?

In summary, two students are standing on a balcony 23.4 m above the street. One student throws a ball down at 15.5 m/s while the other throws a ball up at the same speed. The time difference between the balls in the air is 3.6 seconds and the velocity of each ball when it hits the ground is 0 m/s for the first ball and 30 m/s for the second ball. After 0.480 seconds, the balls are 34.8 meters apart.
  • #1
Leo34005
51
0
Velocity Physics Problem

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
 
Last edited:
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  • #2
Need some help here
 
  • #3
Wouldn't your g = 9.8m/s due to gravity ? As for your first question it would be the ball going down, as you throw the ball up it is going against gravity thus making it achieve its goal much slower than the ball going down.
 
  • #4
I think what i did was wrong, and yours too, so can anyone out there help me find the answer? At least i attempted it
 

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