Question

In a local bar, a customer slides an empty beer mug down the counter for a refill. The height of the counter is 1.30 m. The mug slides off the counter and strikes the floor 0.80 m from the base of the counter.

(a) With what velocity did the mug leave the counter?

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

Answer 1

time of flight in y-direction is given by:

y = Vyi*t + 0.5*ay*t^2

-h = 0*t - 0.5*g*t^2

t = sqrt (2*h/g)

Now distance traveled in x-direction will be:

x = Vxi*t

Since axi = 0

Suppose x = d

then

Vxi = d/sqrt (2*h/g)

So,

Vxi = 1.3/sqrt (2*0.80/9.81) = 3.218m/sec

B.

Now

Vyi = Vxi = 3.218 m/sec

Vyf = Vyi + ay*t

Vyf = 0 - g*t

Vyf = -sqrt (2*g*h)

Vyf = -sqrt (2*9.81*0.80) = -3.961m/sec

So, direction will be:

theta = arctan (Vyf/Vxf)

theta = arctan (3.961/3.218) = -50.88deg

Direction = 50.88deg below the horizontal