A bartender slides a beer mug at 1.4 m/s towards a customer at the end of a frictionless bar that is 1.8 m tall. The customer makes a grab for the mug and misses, and the mug sails off the end of the bar.

(a) How far away from the end of the bar does the mug hit the floor?

(b) What are the speed and direction of the mug at impact?

do vertical problem first to find time in air

1.8 = 4.9 t^2
t = .606 seconds to fall
so it lands
d = 1.4 * .606 = .849 m from bar

horizontal component of impact speed = 1.4 m/s
to find vertical component
v = 0 + 9.8(.606) = 5.34 m/s

speed = sqrt (1.4^2 + 5.34^2)
tan angle down from horizontal = 5.34/1.4

please

(a) Well, that mug sure knows how to make an exit! To find out how far away from the end of the bar it hits the floor, we need to consider the time it takes for it to fall. We can use the equation for vertical motion:

h = vit + (1/2)gt^2

Where h is the height of the bar, vi is the initial vertical velocity, g is the acceleration due to gravity (which is approximately 9.8 m/s^2), and t is the time.

Since the initial vertical velocity is zero (assume the mug is only moving horizontally), we can simplify the equation to:

h = (1/2)gt^2

Plugging in the values, we get:

1.8 m = (1/2)(9.8 m/s^2)t^2

Now, we can solve for t:

t^2 = (2 * 1.8 m) / 9.8 m/s^2

t^2 = 0.367 m

t ≈ 0.606 s

Now that we have the time, we can calculate the horizontal distance the mug covers by multiplying it with the initial horizontal velocity:

d = vit

d = (1.4 m/s)(0.606 s)

d ≈ 0.848 m

So, the mug will hit the floor approximately 0.848 meters away from the end of the bar.

(b) As for the speed and direction of the mug at impact, it's going to hit the floor with a speed that's solely determined by gravity. Since it's only experiencing vertical motion when it falls, the speed at impact will be the same as the free-fall speed for an object falling from a height of 1.8 m. And that speed, my friend, is around 4.24 m/s.

However, since the mug was initially moving horizontally at 1.4 m/s, it will retain that horizontal velocity as it falls. So, the direction of the mug at impact will be purely vertical, while also maintaining its original horizontal velocity.

Just remember, gravity can be a real party pooper sometimes, especially when you're reaching for that beer mug!

To answer these questions, we need to analyze the motion of the beer mug as it slides off the end of the bar.

(a) First, let's find out how long the mug takes to hit the floor. We can use the equation of motion:

h = ut + (1/2)at^2,

where h is the vertical distance covered by the mug, u is the initial velocity vertically downward, a is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time taken.

Since the mug starts from rest vertically, its initial vertical velocity (u) is 0 m/s. The vertical distance (h) is given by the height of the bar, which is 1.8 m.

Therefore, the equation becomes:
1.8 = (1/2)(9.8)t^2.

Rearranging the equation, we get:
t^2 = (2 * 1.8) / 9.8.

Simplifying further, we find:
t^2 = 0.3673.

To find t, we take the square root of both sides:
t = √0.3673.

Calculating this, we get:
t ≈ 0.605 seconds.

Now, to find the horizontal distance covered by the mug before hitting the floor, we can use the formula:

s = ut + (1/2)at^2,

where s is the horizontal distance, u is the initial velocity, a is the acceleration (which is 0 since there is no horizontal force acting), and t is the time taken.

The initial horizontal velocity (u) of the mug is given as 1.4 m/s.

Plugging in the values, we have:
s = (1.4 * 0.605) + (1/2)(0)(0.605)^2.

Simplifying this, we find:
s = 0.845 + 0.

Therefore, the mug hits the floor approximately 0.845 meters away from the end of the bar.

(b) Since there are no horizontal forces acting on the mug, its horizontal velocity does not change. Therefore, its speed at impact remains 1.4 m/s, and its direction depends on the angle at which the mug slides off the bar. It will likely be a downward angle due to the height of the bar, but the exact angle is not provided in the question.