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)
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?
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