Calculating Maximum Height & Vertical Velocity of Kong Throws

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Homework Help Overview

The discussion revolves around a physics experiment involving the motion of a kong toy thrown in the air. Participants are tasked with calculating the maximum height, horizontal velocity, and vertical velocity upon impact based on recorded distances and times for two throws.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between distance, time, and velocity, with one suggesting the use of formulas to derive vertical speed from vertical distance. Another participant proposes using half the total time to find maximum height through a specific kinematic equation.

Discussion Status

Some participants are exploring different methods to calculate the required values, while others express uncertainty about the formulas involved. There is an ongoing exchange of ideas, with attempts to clarify the approach to finding maximum height and vertical velocity.

Contextual Notes

Participants are working with limited information and are trying to derive necessary formulas based on the experiment's parameters. The discussion reflects a mix of understanding and confusion regarding the application of kinematic equations.

Charlie_russo
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(I have already posted this problem in the physics help, but it is more math than physics, so I am also posting it here)

For a physics experiment, we had to take a kong (dog toy with a rubber ball on the end of a rope), twirl it and then throw it as far as we could. We were told to record how long it was in the air for, and the distance for two throws. This was all was we were told to record. Afterwards, we need to find the maximum height the kong reaches, it's horizontal velocity as it is in the air and it's vertical velocity when it hits the ground.

Throw #1-

Distance: 52m
Time: 4.9s

Throw #2-

Distance: 65m
Time: 3.84s



So far, I have been able to figure out the horizontal velocity by using V=d/t.

#1- 52m/4.9s=10.61m/s

#2- 65m/3.84s=16.93m/s

Now, I think that in order to find the vertical velocity, I would need to first find the kongs maximum height, and then divide this by half of the time, (since at the halfway point the kong will now be traveling horizontally, and would take the same amount of time to reach the ground as a kong that is dropped from the same height). However, I am not sure on how to find the maximum height.
 
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Do you know the formulas for the horizontal a vertical distances and speeds? Putting distance and time into the formula for vertical distance should tell you the initial vertical speed and then you can use the distance formula, with half the time, to find the maximum height.
 
no, I don't know the formula, :(
 
ok, i think i figured out how to solve it. at half the total time, the kong will be going horizontal, so the time it will take to reach the ground is the same as if it was just dropped from the same height. I can than use x=1/2(a)(t^2) to find the height. WOuld this work right?
 
ok, here is what i got...

#1-

Horizontal Velocity- 10.61m/s
Vertical Velocity- 12.25m/s
Maximum Height- 30.01m

#2-

Horizontal Velocity- 16.93m/s
Vertical Velocity- 9.6m/s
Maximum Height- 18.43m
 

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