Finding the time of arrival between two balls in air using kinematics

AI Thread Summary
The discussion focuses on solving a kinematics problem involving two balls: one thrown upward with an initial speed of 51 m/s and another dropped from a height of 38 m. The goal is to determine when both balls reach the same height. Participants suggest using kinematic equations to express the height of each ball as a function of time and then equate these expressions to find the time of arrival. There is confusion regarding the use of final velocity (Vf) and the correct approach to set up the equations. Ultimately, the correct method involves equating the height equations for both balls to solve for the time when they are at the same height.
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Homework Statement


A ball is thrown upward from the ground with
an initial speed of 51 m/s; at the same instant,
a ball is dropped from a building 38 m high.
After how long will the balls be at the
same height? The acceleration of gravity is
9.8 m/s2 .
Answer in units of s.


Homework Equations



Equations of Kinematics


The Attempt at a Solution


I thought that by finding the time difference between the two that it would give me the answer.

Setting up the equation as DeltaT = T1 - T2, I found the times for each object using equation Vf = Vi + at for the object with a velocity of 51m/s, and the equation x = Vo*t + 1/2at^2 for the object 38 meters above the ground. Finding the time for these two, I plugged them into my DeltaT equation, with the biggest value as T1.

I got the answer 2.42s, but this was wrong.
 
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hi garcia1! :wink:
garcia1 said:
Setting up the equation as DeltaT = T1 - T2, I found the times for each object using equation Vf = Vi + at for the object with a velocity of 51m/s …

i don't understand … how do you know Vf ? :confused:

why not use the same equation for both balls (with different x0 and v0 of course)? :smile:
 
What I thought you meant by this was that I should set both equations equal to a common variable, and then solve for T accordingly. I tried this with the variable Vf, solving for both in this way:

Eq 1: Vf = Vo +at
Eq 2: Vf^2 = Vo^2 + 2ax -> Vf = rad(Vo^2 + 2ax)

Then setting them equal to each other: Vo + at = rad(Vo^2 + 2ax) - Vo
a

I solved for t, and got the answer T = 2.42s, but this was wrong. Any thoughts on where I messed up?
 
The second equation is over a. Sorry, computing error.
 
You're trying to find when both balls will be at the same height. So wouldn't it be a good idea to write an equation for each ball which gives the height at a given time and then equate them?
 
hi garcia1! :smile:

(just got up :zzz: …)
garcia1 said:
What I thought you meant by this was that I should set both equations equal to a common variable, and then solve for T accordingly. I tried this with the variable Vf …

but that will only give you the time T at which they both have the same velocity, won't it? :wink:
 
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