What is the formula for θ as a function of time in a projectile's path?

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To derive the formula for the angle θ as a function of time for a projectile, the horizontal and vertical components of velocity must be considered. The horizontal velocity remains constant, while the vertical velocity is influenced by gravity. By using the relationship t = x/(Vox cos θ) and substituting this into the vertical position equation, one can express θ in terms of time. The discussion emphasizes that solving for θ requires understanding the projectile's parabolic path and applying trigonometric properties. Overall, the approach involves breaking down the motion into its components and using the derived relationships to find θ.
k-rod AP 2010
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Homework Statement


A ball is thrown horizontally from the top of a cliff with an initial Vo(at t=os). At any moment , its direction of motion makes an angle θ to the horizontal.

1)Derive a formula for θ as a function of time, t, as the ball follows a projectile's path.

Homework Equations


The Attempt at a Solution


components: Vx=Voxcosθ Vy=Voysinθ

then i plugged the x component to find t:Vx=x/t... t=x/(Voxcosθ)

next i plugged t into a y position function Y=Yo+Voyt+1/2at^2

0=h+(Voysinθ )(x/Voxcosθ)+(1/2*-9.8*(x/Voxcosθ)^2)

this be correct because if u were given all of the initial heights and velocities the only variable left to solve for would be θ, giving you your answer right?
 
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That's a pretty long equation. The horizontal velocity is always constant: v_x = v_ox.
The vertical velocity is just v_y = gt.
You can solve for theta as a function of t using basic trig properties of a right triangle.
 
hey thanks, but u can only solve for theta using trig for a 1-d motion problem, but in 2-d the object follows a parabolic path so its velocity must be broken into components.

the t=x/v_o cos(theta) gives you time which you plug into the y position function, which accounts for the parabolic path.

if anybody else knows what to do, any help would be aprreciated
 
Does anybody else know how to solve this, I am really curious what the correct procedure is...
 
If it helps you, I would start by sketching out a graph showing the function of time (Y) and its relationship to t. You can completely ignore X since it will have nothing to do with how quickly Y changes. Y is only a function of time.

From there on out, follow PhanthomJay's suggestions.
 
After re-reading your OP, I'm sorry for replying with an incorrect response. I misunderstood your question (whoops!) :redface:
...its direction of motion makes an angle θ to the horizontal.

Use the equations for your X and Y components of your velocity and work out θ using trig.
 
no worries Condensate =P.. and ok i will try that when it comes around to AP review time, thanks for your help.
 
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