Find angle with given distance and height

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SUMMARY

The discussion focuses on solving for the angle of projectile motion given the horizontal distance (x) and vertical height (y) from a desk. The relevant equations include y = vyot + (½)ayt² and x = vxot, with the relationships vsinΘ = vyo and vcosΘ = vxo. The user attempts to manipulate these equations to isolate Θ but struggles with the algebraic transformations required. A suggestion is made to utilize trigonometric identities to assist in the solution process.

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Homework Statement


You have a device that shoots an object at an angle, on top of a desk.
You have x,y and v solve for the angle.
Origin is at the top of the desk.

Homework Equations


y=vyot+(½)ayt2
x = vxot
vsinΘ = vyo
vcosΘ = vxo

The Attempt at a Solution


y= vsinΘt + (1/2)at^2
x = vcosΘt → t = x/(vcosΘ)

y = x*(vsinΘ)/(vcosΘ) +(1/2)a(x/(vcosΘ))^2
y = xtanΘ + (a/2)*((x^2)/(v^2cos^2Θ))
from here i moved cos^2Θ to the left and changed tanΘ into sincos but I don't know how to isolate the Θ by itsself.

Does anyone have any ideas? I know quadratic equation can be used but i don't know how to change it for Θ

Any help is appreciated.[/B]
 
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Hi bryce:

Your title says you want the angle given the distance and height. Your problem statement does not mention a distance and height, ether as a number or as a variable. I suppose you might have in mind that x is distance and y is height, but then what has the desk got to do with it?

Regards,
Buzz
 

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