Question

In a local bar, a customer slides an empty beer mug down the counter for a refill. The height of the counter is 1.44 m. The mug slides off the counter and strikes the floor 1.60 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?​ ° (below the horizontal)​

Answer 1

Say the horizontal component of the velocity is vx and the vertical is vy.

Initially at t=0 (as the mug leaves the bar) the components are v0x and v0y.

Obviously v0y = 0.

The equations for horizontal and vertical projectile motion (with the positive direction up) are

x = x0 + v0x t

y = y0 + v0y t - 1/2 g t^2 = y0 - 1/2 g t^2

Now choose the origin to be the end of the counter. x0=0 and y0=0. The equations simplify to

x = v0x t

y = - 1/2 g t^2

You know that x = 1.6m when y = -1.44m

From the y equation (and g=9.8 m/s^2) you can calculate the time that the mug hits the floor.

t = 0.5421s

From the x equation we get the initial horizontal velocity

v0x = x/t = 1.6/0.5421 = 2.95 m/s

(b) x-component of velocity is constant since there are no horizontal forces so vx = 2.95 m/s

y-component is given by v = u+at with u=0 and a=-g

vy = -gt = -5.31 m/s

Now tan(angle) = vy/vx so angle = arctan(vy/vx)

= arctan (5.31/2.95) = 60.96 degrees