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 a distance d from the base of the counter. if the height of the counter is h. a) find an expression for the time t it takes the cup to fall to the floor in terms of the variables h and g. b) with what speed does the mug leave the counter? answer in terms of variables d,g, and h. c) in the same terms what is the speed of the cup immediately before it hits the floor? d) in terms of h and d what is the direction of the cups velocity immediately before it hits the floor?

Question

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 a distance d from the base of the counter. if the height of the counter is h. a) find an expression for the time t it takes the cup to fall to the floor in terms of the variables h and g. b) with what speed does the mug leave the counter? answer in terms of variables d,g, and h. c) in the same terms what is the speed of the cup immediately before it hits the floor? d) in terms of h and d what is the direction of the cups velocity immediately before it hits the floor?

Answer 1

a) To find the time it takes for the cup to fall to the floor, we'll use the equation for free-fall motion: h = 0.5 * g * t^2. Solving for time t, we get:

t^2 = 2 * h / g
t = sqrt(2 * h / g)

b) Since the cup slides off the counter horizontally, its horizontal velocity does not change during the fall (there's no friction or air resistance). We can use the distance formula relating speed, time, and distance to get the horizontal speed (v_x) of the mug leaving the counter: d = v_x * t. Solving for v_x, we get:

v_x = d / t
v_x = d / sqrt(2 * h / g)

c) The cup's final speed is the vector sum of its horizontal and vertical components. The vertical component (v_y) can be found using the equation for free-fall motion: v_y = g * t. Plugging in our expression for t, we get:

v_y = g * sqrt(2 * h / g)

Now, to find the total speed (v) immediately before the cup hits the floor, we'll use the Pythagorean theorem:

v^2 = v_x^2 + v_y^2
v = sqrt(v_x^2 + v_y^2)
v = sqrt((d / sqrt(2 * h / g))^2 + (g * sqrt(2 * h / g))^2)

d) The direction of the cup's velocity immediately before it hits the floor can be represented by an angle with respect to the horizontal. We can use the arctangent function to find this angle:

angle = arctan(v_y / v_x)
angle = arctan((g * sqrt(2 * h / g)) / (d / sqrt(2 * h / g)))

Lastly, simplifying the expression, we get:

angle = arctan(g * d / 2 * h)

Answer 2

a) To find the expression for the time it takes the cup to fall to the floor, we can use the equation of motion for free-fall:

h = (1/2) * g * t^2

Where:
h = height of the counter
g = acceleration due to gravity (approximately 9.8 m/s^2)
t = time

Rearranging the equation for t, we get:

t = √(2h/g)

So, the expression for the time it takes the cup to fall to the floor is t = √(2h/g).

b) To find the speed at which the cup leaves the counter, we can use the equation of motion for free-fall:

v = g * t

Substituting the value of t from the previous step, we get:

v = g * √(2h/g)

Simplifying this expression, we get:

v = √(2g * h)

So, the speed at which the cup leaves the counter is v = √(2g * h).

c) The speed of the cup immediately before it hits the floor can be found using the equation of motion for free-fall:

v^2 = u^2 + 2g * d

Where:
v = final velocity
u = initial velocity (which is 0 in this case, as the cup slides off the counter)
g = acceleration due to gravity
d = distance at which the cup strikes the floor

Substituting the values, we have:

v^2 = 0 + 2g * d
v^2 = 2g * d

Taking the square root of both sides, we get:

v = √(2g * d)

So, the speed of the cup immediately before it hits the floor is v = √(2g * d).

d) The direction of the cup's velocity immediately before it hits the floor is downward, as it is falling due to the force of gravity.

Answer 3

a) To find an expression for the time it takes the cup to fall to the floor, we can use the equations of motion. The key concept here is that the horizontal motion (sliding off the counter) does not affect the vertical motion (falling).

Let's assume the cup slides off the counter with an initial horizontal velocity of zero (since it was simply dropped). The only force acting on the cup vertically is gravity (acceleration due to gravity, g). Thus, the equation for the vertical motion can be written as:

h = (1/2) * g * t^2

Where:
h = height of the counter
g = acceleration due to gravity (approximately 9.8 m/s^2)
t = time it takes the cup to fall

We are trying to find an expression for t, so let's rearrange the equation:

t^2 = (2 * h) / g

Taking the square root of both sides:

t = sqrt((2 * h) / g)

b) To find the speed at which the mug leaves the counter, we can use the equation for the horizontal motion. Since the cup slides off the counter with an initial horizontal velocity of zero:

d = 0 * t + (1/2) * 0 * t^2

d = 0

Therefore, the cup leaves the counter with zero horizontal speed.

c) The speed of an object just before it hits the ground is equal to the speed it has gained during free fall. So, in terms of variables d, g, and h, the speed of the cup just before it hits the floor is given by the equation:

v = sqrt(2 * g * h)

d) Since the cup is falling vertically, its velocity is directed downwards. Thus, the direction of the cup's velocity immediately before it hits the floor is downwards.

In summary:
a) The expression for the time it takes the cup to fall to the floor is t = sqrt((2 * h) / g).
b) The mug leaves the counter with zero horizontal speed.
c) The speed of the cup just before it hits the floor is v = sqrt(2 * g * h).
d) The direction of the cup's velocity immediately before it hits the floor is downwards.