In a local bar, a customer slides an empty beer mug down the counter for a refil

Question

In a local bar, a customer slides an empty beer mug down the counter for a refill. The bartender is momentarily distracted and does not see the mug, which slides off the counter and strikes the floor at distance d from the base of the counter. The height of the counter is h. (Use any variable or symbol stated above along with the following as necessary: g.)
(a) With what speed did the mug leave the counter?



v

=






(b) What was the direction of the mug's velocity just before it hit the floor?



?

=




radians below horizontal

Explanation / Answer

Let v be the velocity of the mug when it just leave the counter. Since there isnt any opposing force along the x-direction, the time it will take to travel a distance 'd' horizontally is given by t = d / v. In this same time, the mug has travelled a vertical distance of 'h'. Its initial vertical velocity is zero and under the gravitational pull hence in this time, the distance travelled is given by h = 0 + (1/2) * g * t^2 = (g/2) (d/v)^2. => v = d / sqrt(2h/g) At this time, the vertical velocity = vy = u + at = 0 + gt = gt = g (d/v) = g * d / d / sqrt(2h/g) = g * sqrt(2h/g) = sqrt(2gh). and the horizontal velocity is the same as initial velocity. Hence vx = v = d / sqrt(2h/g) Hence the direction of mugs velocity now = ? = arctan(vy / vx) = arctan[ sqrt(2gh) / d / sqrt(2h/g)] = arctan(2h/d)

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