Calculate Velocity from Cliff Jump: Horizontal and Vertical Distance

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SUMMARY

The discussion focuses on calculating the velocity of a person jumping off a cliff at a 26-degree angle, covering both horizontal and vertical distances of 23 meters and 12 meters, respectively. The relevant equations used are X=Vxt + t for horizontal motion and Y=Vyt + 1/2at² for vertical motion. The participant expresses confusion regarding the differences between this scenario and a standard 45-degree projectile problem. The thread also references additional resources for understanding trajectory problems.

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Lordrunt
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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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