Question

In a local bar, a customer slides an empty beer mug down the counter for a refill. The height of the counter is 1.10 m. The mug slides off the counter and strikes the floor 0.20 m from the base of the counter.
(a) With what velocity did the mug leave the counter?
____________m/s ????

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

Answer 1

horizontal component of the velocity is vx and the vertical is vy.
Initially at t=0 the components are v0x and v0y.
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 = 1.10m when y = -0.20m
From the y equation (and g=9.8 m/s^2) you can calculate the time that the mug hits the floor.
t = 0.2020s
From the x equation we get the initial horizontal velocity
v0x = x/t = 1.1/0.20 = 5.5 m/s

(b) x-component of velocity is constant since there are no horizontal forces so vx = 5.5 m/s
y-component is given by v = u+at with u=0 and a=-g
vy = -gt = -9.8*0.20m/s = 1.96 m/s

magnitude of velocity = V = 5.83 m/s

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

arctan(1.96/5.5)= 19.61 degree