How long do the balls take to hit the ground?

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Homework Help Overview

The problem involves two balls thrown from a balcony, one downward and one upward, with a focus on their time in the air, final velocities, and separation distance after a specific time. The subject area is kinematics in physics.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the motion equations used for both balls, with one student attempting to calculate the time of flight and the difference in time spent in the air. Questions arise regarding the acceleration due to gravity and its impact on the calculations.

Discussion Status

Some participants are questioning the initial assumptions made about gravity and the calculations presented. There is an acknowledgment of attempts made, but also a recognition that there may be errors in the reasoning or calculations. Guidance is sought to clarify these points.

Contextual Notes

There is a mention of differing values for gravitational acceleration, with one participant suggesting 9.8 m/s² instead of 10 m/s². The original poster expresses uncertainty about their approach and seeks further assistance.

Leo34005
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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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Need some help here
 
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.
 
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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