A ball launched on a imclined plane

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SUMMARY

The discussion centers on the physics problem of a ball launched from a cart rolling down an inclined plane at an angle θ. The key equations involved are yf = yi + Viy*t + 1/2ay*t^2 and xf = xf + Vix*t + 1/2ax*t^2. The main question is whether the ball lands in the funnel, to the right, or to the left of it upon returning to the same height. The participant is confused about the launch direction of the ball and the relationship between the velocities of the ball and the cart.

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This discussion is beneficial for physics students, educators, and anyone interested in understanding the dynamics of motion on inclined planes and projectile trajectories.

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Homework Statement


The cart is rolling down an inclined plane which is at an angle  to the horizontal.
When the cart’s speed is V, the ball is launched. When the ball returns to the same initial perpendicular distance above the track, does it land in the funnel, to the right of the funnel, or to the left of the funnel?


Homework Equations


yf = yi + Viy*t + 1/2ay*t^2
xf = xf + Vix*t + 1/2ax*t^2

The Attempt at a Solution


I was thinking then the Xball would be something like
Xf ball = Vix(ball)*t + V(launch x) * t
Xf car = Vix(car)*t + 1/2(gx)t^2
Vix ball / car are equal...so in this case i have to determine whether V(launch x) * t is larger or 1/2(gx)t^2 is larger...there were no numbers given so i and very confused right now thinking maybe my attempt is on the wrong track, can someone help me?
 
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Is the ball launched perpendicular to the ground(x axis) or to the cart(inclined plane)?
 
perpendicular to the cart
 

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