Question

PROJECTILE MOTION

A bartender slides a beer mug at 1.50 m/s towards a customer at the end of a frictionless bar that is 1.20 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?

Answer 1

I've answered Part A, and I got 0.75 m.

Answer 2

I'll show the work in a minute or two.

Answer 3

the velocity has two components:
1) the horizontal component whose magnitude is 1.50 m/s
2) the vertical component whose magnitude is:
\[\sqrt {2gh} = \sqrt {2 \cdot 9.81 \cdot 1.2} = ...?\]

Answer 4

I first found \(t\).
\[\Delta y=V_0y t+ \frac{1}{2}at^2\]\[-1.2= -4.9t^2\]\(t=0.5s\)


Then I found for \(\Delta x\)/.
\[\Delta x=V_0xt\]\[\Delta x=(1.50)(0.5)\]\[\Delta x=0.75m\]

Answer 5

Wait, so should I find the y component first using the formula you showed or that's for the velocity?

Answer 6

since your mug, at impact, gains a vertical speed

Answer 7

And I got 5.077 m/s.

Answer 8

or 5.08 m/s

Answer 9

I got a different result:
\[\sqrt {2gh} = \sqrt {2 \cdot 9.81 \cdot 1.2} = 4.85\]

Answer 10

do you agree?

Answer 11

No, because that's just the y component, and what I did it I found the y component (4.85) and the x component (1.5), then I used \(v=\sqrt {(4.85)^2+(1.5)^2}\).

Answer 12

ok! your answer is correct, I meant the vertical speed is 4.85

Answer 13

Ohh I was thinking ahead. Then to find the direction, should l just use tan^-1 V_fy/V_fx?

Answer 14

which gives me 72.81 degrees.

Answer 15

correct!
for direction we can write this:
\[4.85 = 1.5 \cdot \tan \theta \]

Answer 16

Hold on, it would be negative, right?

Answer 17

no, since we consider this angle
\(\beta=180-72.81=...\)