Finding Energy while knowing intial Y and desired x

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SUMMARY

The discussion focuses on calculating the energy required for an object launched from a height of 1.7526 meters to reach a horizontal distance of 1.8288 meters. The user initially calculated the hypotenuse of the trajectory but was advised that this approach is incorrect. Instead, the correct method involves using the standard constant acceleration equations in both the x and y directions, taking into account the initial speed (v) and launch angle (θ).

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Homework Statement



An object is launched at x m/s from a 1.7526 meter platform. How much energy must be exerted for the object to be launched 1.8288 meters?

Homework Equations



Not Sure

The Attempt at a Solution



I made a right triangle and found that the hypotenuse (or distance that the object will need to travel) is 2.532888 meters.
 
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welcome to pf!

hi sfishman59! welcome to pf! :wink:

do you mean that the object is launched at an unknown angle from 1.7526 m above the ground, and has to land on the ground 1.8288 m away?

the hypotenuse has nothing to do with it

call the initial speed v and the initial angle θ, and use the standard constant acceleration equations in the x and y directions (separately, with a = 0 and a = -g) …

show us what you get :smile:
 

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