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Physics For Scientists & Engineers
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Short Answer

In a local bar, a customer slides an empty beer mug down the counter for a refill. The height of the counter is h. The mug slides off the counter and strikes the floor at distance d from the base of the counter.

(a) With what velocity did the mug leave the counter?

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

  1. The velocity of the mug when it leaves the counter is d·g2·h.
  2. The direction of the mug’s velocity just before it hit the floor is 2·hd.
See the step by step solution

Step by Step Solution

Step 1: Given Data

  • The height of the counter is h.
  • The mug slides off the counter and strikes the floor at distance d from the base of the counter.

Step 2: Definition of velocity

The rate of change of an object's location with regard to a frame of reference is its velocity, which is a function of time. A statement of an object's speed and direction of motion is referred to as velocity.

Step3: Find the velocity of the mug when it leaves the counter.

(a)

The initial angle is θ0=0 and the initial height is y0=h.

Therefore,

x=v0·ty=h-12·g·t2

The maximum distance is xmax=d and the final height is y=0:

d=v0·tmax0=h-12·g·tmax2

Rewrite the equations above:

v0=dtmaxtmax=2·hg

Substitute the second equation into first and calculate the result:

v0=d2·hg=d·g2·h

Step 4: Find the direction of the mug’s velocity just before it hit the floor. 

(b)

The vx component of motion is constant in horizontal projectile motion, but the vy component rises in magnitude:

width="94" style="max-width: none; vertical-align: -41px;" vx=v0vy=2·g·y

The mug's velocity components at the instant of collision with the ground are:

vx=v0=d·g2·hvy=2·g·h

The velocity angle θ at the instant of contact may be computed as follows:

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