A physics student playing with an air hockey table (a frictionless surface) find

Question

A physics student playing with an air hockey table (a frictionless surface) finds that if she gives the puck a velocity of 3.86m/s along the length ( 1.82m ) of the table at one end, by the time it has reached the other end the puck has drifted a distance 2.51cm to the right but still has a velocity component along the length of 3.86m/s . She concludes correctly that the table is not level and correctly calculates its inclination from the above information.
What is the angle of inclination?

I would like a step by step explanation please. Thank ypu

Explanation / Answer

Let: t be the time the puck is moving, a be the tilt from horizontal. Along the table: 1.68 = 3.77t ...(1) Across the table, the acceleration is the component of g parallel to the surface of the table, namely g sin(a), as the perpendicular component g cos(a) is offset be the tilted air jets. Applying the equation for motion with uniform acceleration g sin(a): 2.40 = 9.80 t^2 sin(a) / 2 ...(2) Substituting for t from (1) in (2): 2.40 = 9.80 (1.68 / 3.77)^2 sin(a) / 2 sin(a) = 2 * 2.40 / [ 9.80(1.68 / 3.77)^2 ] = 2.47. That's impossible. Have you noticed the sideways 'drift' is greater than the length of the table? Surely that means the puck ends up on the floor! If the puck were falling vertically, in the time available it would fall: s = 9.80(1.68 / 3.77)^2 / 2 = 0.97. It cannot possibly drift 2.40 to the right.

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