Physics, 18.02.2020 23:53 JANA279

In a local diner, a customer slides an empty coffee cup down the counter for a refill. The cup slides off the counter and strikes the floor at distance d from the base of the counter.
(a) If the height of the counter is h, find an expression for the time t it takes the cup to fall to the floor in terms of the variables h and g.
t =
(b) With what speed does the mug leave the counter? Answer in terms of the variables d, g, and h.
v1 =
(c) In the same terms, what is the speed of the cup immediately before it hits the floor?
v2 =
(d) In terms of h and d, what is the direction of the cup's velocity immediately before it hits the floor?
? = radians below horizontal

Answers: 2

Show answers

Answers

Answer from: zariah10191101

a) $t=\sqrt{\frac{2h}{g}}$

b) $v=\frac{d}{\sqrt{\frac{2h}{g}}}$

c) $v=\sqrt{d^2(\frac{g}{2h})+(2gh)}$

d) $\theta=tan^{-1}(\frac{2h}{d})$ (radians)

Explanation:

The motion of the cup sliding off the counter is the motion of a projectile, consisting of two independent motions:

- A uniform motion along the horizontal direction

- A uniformly accelerated motion (free fall) along the vertical direction

The time of flight of the cup is entirely determined by the vertical motion, therefore we can use the suvat equation:

$s=ut+\frac{1}{2}at^2$

where here:

s=h (the vertical displacement is the height of the counter)

u=0 (the initial vertical velocity of the cup is zero)

a=g (the vertical acceleration is the acceleration of gravity)

Solving for t, we find the time of flight of the cup:

$h=0+\frac{1}{2}gt^2\\t=\sqrt{\frac{2h}{g}}$

To solve this part, we just analyze the horizontal motion of the cup.

Here we know that the horizontal motion of the cup is uniform: this means that is horizontal speed is constant during the whole motion, and it is actually equal to the speed at which the mug leaves the counter.

For a uniform motion, the speed is given by

$v=\frac{d}{t}$

where

d is the distance covered

t is the time taken

Here, the distance covered is , the distance from the base of the counter, while the time taken is the time of flight:

$t=\sqrt{\frac{2h}{g}}$

Substituting into the previous equation, we find the speed of the mug as it leaves the counter:

$v=\frac{d}{\sqrt{\frac{2h}{g}}}$

Here we want to find the speed of the cup immediately before it hits the floor.

Here we have to consider that while the mug falls, its vertical speed increases, while the horizontal speed remains constant.

Therefore, the horizontal speed of the cup before it hits the ground is:

$v_x=\frac{d}{\sqrt{\frac{2h}{g}}}=d\sqrt{\frac{g}{2h}}$

The vertical speed instead is given by the suvat equation:

v_y=u_y + at

where:

u_y=0 is the initial vertical velocity

a=g is the acceleration

$t=\sqrt{\frac{2h}{g}}$ is the time of flight

Substituting,

$v_y = 0 +g(\sqrt{\frac{2h}{g}})=\sqrt{2gh}$

The actual speed of the cup just before it hits the floor is the resultant of the horizontal and vertical speeds, so it is:

$v=\sqrt{v_x^2+v_y^2}=\sqrt{d^2(\frac{g}{2h})+(2gh)}$

Just before hitting the floor, the velocity of the cup has two components:

$v_x=d\sqrt{\frac{g}{2h}}$ is the horizontal component (in the forward direction)

$v_y=\sqrt{2gh}$ is the vertical component (in the downward direction)

Since the two components are perpendicular to each other, the angle of the direction is given by the equation

$tan \theta = \frac{v_y}{v_x}$

where here $\theta$ is measured as below the horizontal direction.

Substituting the expressions for v_x,v_y , we find:

$tan \theta = \frac{\sqrt{2gh}}{d\sqrt{\frac{g}{2h}}}=\frac{2h}{d}$

$\theta=tan^{-1}(\frac{2h}{d})$ (radians)

Answer from: Quest

atomic = no of protons.

mass number-atomic number = neutron number

Answer from: Quest

a electromagnetic waves

Another question on Physics

Physics, 22.06.2019 03:00

If a spring has a k value of 100 newtons per meter and it is stretched 0.50 meters, what is the restoring force of the spring?

Answers: 2

Answer

Physics, 22.06.2019 03:30

Will give brainliest! jay rides his 2.0-kg skateboard. he is moving at speed 5.8 m/s when he pushes off the board and continues to move forward in the air at 5.4 m/s. the board now goes forward at 13 m/s.a. determine jay’s mass.b. determine the change in the internal energy of the system during this process.(express your answer to two significant figures and include the appropriate units.)

Answers: 1

Answer

Physics, 22.06.2019 05:00

Aperson walking 1 mile everyday for exercise, leaving her front porch at 9: 00 am and returning to her front porch at 9: 25 am. what is the total displacement of her daily walk? a) 1 mileb) 0c) 25 minutes d) none of the above

Answers: 1

Answer

Physics, 22.06.2019 19:40

It may seem strange that the selected velocity does not depend on either the mass or the charge of the particle. (for example, would the velocity of a neutral particle be selected by passage through this device? ) the explanation of this is that the mass and the charge control the resolution of the device--particles with the wrong velocity will be accelerated away from the straight line and will not pass through the exit slit. if the acceleration depends strongly on the velocity, then particles with just slightly wrong velocities will feel a substantial transverse acceleration and will not exit the selector. because the acc

Answers: 1

Answer

You know the right answer?

In a local diner, a customer slides an empty coffee cup down the counter for a refill. The cup slide...

Questions