Calculate Velocity from Cliff Jump: Horizontal and Vertical Distance

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AI Thread Summary
To calculate the velocity of a person jumping off a cliff at a 26-degree angle, one must consider both horizontal and vertical distances traveled: 23 meters horizontally and 12 meters vertically. The equations of motion, X=Vxt + t and Y=Vyt + 1/2at^2, are essential for solving the problem. The challenge arises in understanding how to apply these equations differently compared to a 45-degree jump, as the angle affects the components of velocity. Clarification on the techniques for solving trajectory problems is available in related forum discussions. Mastering these concepts is crucial for accurately determining the velocity in this scenario.
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Homework Statement


Example: A person jumped off of a cliff into the water at a 26 degree angle. He traveled 23 meters (horizontally) and fell 12 meters (vertically). What was the person's velocity?


Homework Equations


X=Vxt + t
Y=Vyt + 1/2at^2



The Attempt at a Solution


I was stuck instantly and I was sick the day we took these notes. So I don't know how to do these types of problems. (unless they're the 45 degree one)
 
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How is this so much different than a 45 degree problem? The only difference is the angle, wouldn't you just be able to do it like a 45 degree problem but use 26 degrees instead of 45?
 
Lordrunt said:

Homework Statement


Example: A person jumped off of a cliff into the water at a 26 degree angle. He traveled 23 meters (horizontally) and fell 12 meters (vertically). What was the person's velocity?


Homework Equations


X=Vxt + t
Y=Vyt + 1/2at^2



The Attempt at a Solution


I was stuck instantly and I was sick the day we took these notes. So I don't know how to do these types of problems. (unless they're the 45 degree one)

Please see my posts in this similar thread about trajectory problems. They should help you understand the techniques used for these types of problems:

https://www.physicsforums.com/showthread.php?t=655518

.
 
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