physics people..just a question?

Question

A ball is being released onto a ramp that is on a table that ends where the ramp does. When the ramp stops, the ball will fly into the air into a can.

Using this equation: ∆y = vy i t + ½ gt2 (the gt is raised to the 2 power, not times two), what is the time it would take, t, for a ball to fall from the bottom end of the ramp and land into a can that is 10.5 cm tall

g= -9.81m/s
vx = .63m
the distance from the ramp to the can is 1.75 m

height of table= 1 meter
vyi = 0

Answer 1

Is 10.5 cm tall significant? Are you supposed to get the time the ball passes the top of the can? I'll assume yes. The ball must fall 1 - 0.105 m in the time it travels horizontally to the can's horizontal location.
The hor. distance from the ramp edge to the bottom of the can is sqrt(1.75^2 - 1^2) = 1.436 m. If vx is the hor. velocity at release, and the hor. distance is 1.436 m, the value of t should be 1.436/.63 = 2.280 s. If this value conflicts with the value computed from vertical motion we have a badly posed problem.
BTW, g should be in m/s^2 and vx in m/s.
To solve for t based on vertical motion, construct and solve the quadratic .5*g*t^2 + vyi*t - deltay = 0. Since vyi=0 we don't need the discriminant-based quadratic solution. We have t = sqrt(2*deltay/g). Deltay is 1 m minus 0.105 m = 0.895 m. I get 0.427 s for this fall distance. The answer conflicts with the horizontal time of 2.28 s, so something is fishy. Please check the problem and make sure you have described it accurately. Is there a slope on the ramp that increases vx?