Projectile Motion From Cliff Homework: Solving for Distance and Time

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The discussion focuses on solving a projectile motion problem involving a launch from a cliff. The key equations provided include the horizontal and vertical components of velocity, as well as the equations for horizontal and vertical displacement. The main challenge is determining the time of flight (t) to calculate the horizontal distance (delta x) from the cliff's base. It is emphasized that the vertical final velocity (vfy) cannot be assumed to be zero just before impact, and the correct approach involves solving for t using the equation Q = v0sinθ*t - (gt^2)/2. The final solution can be obtained by substituting t back into the equation for delta x.
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Homework Statement



A projectile is launched with speed v0 making angle θ0 with the horizontal from the edge of a vertical cliff at height Q above a horizontal plane. At what distance from the bottom of the cliff does the projectile land?


Homework Equations



v0x=v0cosθ0
v0y=v0sinθ0
deltax=v0sinθ0*t
deltay=v0sinθ0*t
Q=v0sinθ*t-(gt^2)/2

equations that may work?
(v0)^2=vf^2+2ax... ?


The Attempt at a Solution



other than the equations above, I have not been able to solve anything. Basically, I need t to find delta x (the distance from the base of the clifff) and I know that i need to find t in terms of y (but that t in x and y are equal, but motion in x and y is independent)... according to my TA, i cannot assume that vfy=0 and that i must assume that it is just before it hits the ground. How do I find vfy with this assumption (as doing it with this assumption is easy)

Thanks! I have a final on monday 0_o
 
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deltay=v0sinθ0*t is not correct; need to add g*t²
but I don't think you need this anyway.

Just solve your Q=v0sinθ*t-(gt^2)/2 for t.
Sub in deltax=v0sinθ0*t to get the final answer.
 

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