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 height of the counter is h. The mug slides off the counter and strikes the floor at distance d from the base of the counter. (a) With what velocity did the mug leave the counter? (Use any variable or symbol stated above along with the following as necessary: g.) v_xi = (b) What was the direction of the mug's velocity just before it hit the floor? (Use any variable or symbol stated above as necessary.) theta = below the horizontal

Explanation / Answer

Solution- When the mug just leave the counter let v be the velocity.

As there is not any opposing force along x-direction,the time taken by it to travel a

distance d horizontally=t=d/v

Also during this same time,the mug has travelled vertical distance oh h

its initial velocity=0and under gravitational pull

h = 0.5 g t^2

t = sqrt (2h / g)

So the speed of the mug when it left the counter

Vx = d / sqrt (2h / g)

The vertical component of this velocity is:

Vy = g sqrt (2h / g) = sqrt (2 g h)
And horizontal velocity is the same as initial velocity.
So vx = v = d / sqrt(2h/g)
Therefore the direction of mugs velocity now is

= ? = arctan(vy / vx)
= arctan[ sqrt(2gh) / d / sqrt(2h/g)]
= arctan(2h/d)

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