In a local bar, a customer slides an empty beer mug down the counter for a refill. the height of the counter is 1.14 m. the mug slides off the counter and strikes the floor 0.60 m from the base of the counter.

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 (I hope!) 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 = 1.20m when y = -0.88m
From the y equation (and g=10 m/s^2) you can calculate the time that the mug hits the floor.
t = 0.420s
From the x equation we get the initial horizontal velocity
v0x = x/t = 1.2/0.42 = 2.86 m/s

(b) x-component of velocity is constant since there are no horizontal forces so vx = 2.86 m/s
y-component is given by v = u+at with u=0 and a=-g
vy = -gt = -4.2m/s

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

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