A penny on its side moving at speed v slides off the horizontal surface of a table a vertical distance y from the floor.
a) Show that the penny lands a distance v√(2y/g) from the base of the coffee table.
b) If the speed is 3.5 m/s and the coffee table is 0.4 m tall, show that the distance the coin lands from the base of the table is 1.0 m. (Use g = 9.8 m/s^2)
time in air:
y=1/2 g t^2
t=sqrt(2y/g)
horizonal distance=vt=vsqrt(2y/g)
To solve this problem, we can analyze the motion of the penny in two parts: the horizontal motion and the vertical motion.
a) Let's consider the horizontal motion first. Since there are no horizontal forces acting on the penny, its horizontal velocity remains constant. Let's denote the horizontal distance the penny travels as x.
We know that the time taken to travel this horizontal distance is the same as the time taken for the penny to fall vertically from a height y.
Using the equations of motion, we have:
x = v * t (1)
y = (1/2) * g * t^2 (2)
From equation (1), we can rearrange it to express the time taken in terms of x and v:
t = x / v
Plugging this back into equation (2), we get:
y = (1/2) * g * (x / v)^2
y = (g / 2v^2) * x^2 (3)
Now, let's move on to the vertical motion. The penny falls vertically from a height y, so we can use the equation of motion for free fall:
y = (1/2) * g * t^2
Rearranging this equation, we get:
t^2 = (2y) / g
Substituting for t^2 in equation (3), we have:
y = (g / 2v^2) * x^2
y = (g / 2v^2) * (v^2 * t^2)
y = (g / 2v^2) * (v^2 * (2y) / g)
y = (v^2 * 2y) / (2v^2)
y = y
Now, let's find out the horizontal distance the penny travels before reaching the floor. Using equation (1), we can solve for x:
x = v * t
x = v * (x / v)
x = x
Therefore, the penny lands at the same horizontal distance, x, from the base of the coffee table as it was initially placed.
b) Now, let's substitute the given values into the equation we derived to calculate the horizontal distance.
Using g = 9.8 m/s^2, v = 3.5 m/s, and y = 0.4 m, we have:
x = v * √((2y) / g)
x = 3.5 * √((2 * 0.4) / 9.8)
x ≈ 3.5 * √0.0816
x ≈ 3.5 * 0.2857
x ≈ 1.0 m
Therefore, the distance the penny lands from the base of the table is approximately 1.0 m.
To solve this problem, we can use basic kinematic equations.
a) Let's consider the vertical motion of the penny. The initial vertical velocity, u, is 0 m/s since the penny is sliding horizontally. The final vertical velocity, v_f, can be found using the equation:
v_f^2 = u^2 + 2aΔy
Since the penny is only subject to the acceleration due to gravity, which is approximately 9.8 m/s^2, we can simplify this equation to:
v_f^2 = 2gΔy
Now, we need to find the time, t, it takes for the penny to fall from the table to the floor. We can use the equation:
Δy = uyt + (1/2)at^2
Since the initial vertical velocity, u, is 0 m/s and the initial position, y, is the height of the table, we have:
Δy = yt + (1/2)gt^2
Simplifying this equation, we get:
y = (1/2)gt^2
Now, we can substitute this into the equation for v_f:
v_f^2 = 2g[(1/2)gt^2]
v_f^2 = g^2t^2
Taking the square root of both sides, we get:
v_f = gt
Now, we can substitute this expression for v_f into the equation v_f = u + gt:
gt = gt
This shows that the time it takes for the penny to fall is the same as the time it takes for it to reach the ground horizontally. Therefore, the horizontal distance the penny travels, x, is given by:
x = vt
Substituting v = gt, we get:
x = gty
Next, we need to find the time, t, in terms of v and y. From the equation v_f = gt, we can solve for t:
t = v_f / g
Since v_f = gt, we have:
t = v / g
Now, substituting this value into the equation for x, we get:
x = (g / g) * y * (v / g)
Simplifying this expression, we have:
x = v * y / g
Finally, substituting the given values of v = 3.5 m/s, y = 0.4 m, and g = 9.8 m/s^2, we can calculate the distance the penny lands from the base of the table:
x = (3.5 m/s) * (0.4 m) / (9.8 m/s^2)
x = 0.143 m
Therefore, the distance the penny lands from the base of the coffee table is approximately 0.143 m.
b) Substituting the given values into the general expression derived in part a:
x = v * √(2y / g)
x = (3.5 m/s) * √(2 * 0.4 m / 9.8 m/s^2)
x = (3.5 m/s) * √(0.0816)
x ≈ 1.0 m
Therefore, the distance the penny lands from the base of the table is approximately 1.0 m.