Question

In a local bar, a customer slides an empty beer mug down the counter for a refill. The height of the counter is 1.20 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?
m/s

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

Answer 1

given,

height of counter = 1.20 m

horizontal distance from base = 1.6 m

a) velocity of mug = ?

using equation of motion

`s_y = u t + (1)/(2)gt^2`

`1.2 = (1)/(2)* 9.8 * t^2`

`t^2 = 0.245`

t = 0.495 s

speed of the mug

s_x = v x t

1.6 = v x 0.495

v = 3.23 m/s

b) final velocity of mug in y direction

again using equation of motion

v² = u² + 2 a s

v²= 0 + 2 x 9.8 x 1.2

v = √23.52

v_y = 4.85 m/s

now, direction

`\theta = tan^(-1)((v_y)/(v_x))`

`\theta = tan^(-1)((4.85)/(3.23))`

`\theta =56.34^0`