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

Answer

given,

height of counter = 1.20 m

horizontal distance from base = 1.6 m

a) velocity of mug = ?

   using equation of motion

   $s_y = u t + \dfrac{1}{2}gt^2$

   $1.2 = \dfrac{1}{2}\times 9.8 \times t^2$

   $t^2 = 0.245$

     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

  $\theta = tan^{-1}(\dfrac{v_y}{v_x})$

  $\theta = tan^{-1}(\dfrac{4.85}{3.23})$

  $\theta =56.34^0$