In a local bar, a customer slides an empty beer mug down the counter for a refill. The height of the counter is 1.18 m. The mug slides off the counter and strikes the floor 1.20 m from the base of the counter. With what velocity did the mug leave the counter? What was the direction of the mug's velocity just before it hit the floor?

Question

In a local bar, a customer slides an empty beer mug down the counter for a refill. The height of the counter is 1.18 m. The mug slides off the counter and strikes the floor 1.20 m from the base of the counter.  With what velocity did the mug leave the counter? What was the direction of the mug's velocity just before it hit the floor?

ujjwal · Answer

|dw:1347919468881:dw|

ujjwal · Answer

Here $$H=\frac{1}{2}gt^2$$where H is height and t is time of flight 
you will get t from above

Now use the relation R=ut
where R is horizontal range and u is initial velocity (the velocity with which mug leaves counter)

|dw:1347920232001:dw|If Vy be final vertical velocity and Vx be final horizontal velocity
Vy=gt
and Vx=u (horizontal velocity remains constant throughout the motion)

Now by trigonometry$$\theta=\tan^{-1}(\frac{V_y}{Vx})$$