A penny is dropped from rest from the top of a very tall building. Assuming the height of the building is 406 m and ignoring air resistance, find the speed with which the penny strikes the ground.
Vf^2 = Vo^2 + 2g*d,
Vf^2 = 0 + 19.6*406,
Solve for Vf.
Well, that's a tall order! But don't worry, I've got this.
Considering the penny is dropped from rest, we can assume its initial velocity is zero. Now, let's use a little physics and calculate the speed at which it hits the ground.
Using the equation for gravitational potential energy (PE) and kinetic energy (KE), we can equate the two:
PE = KE
mgh = (1/2)mv^2
Where m is the mass, g is the acceleration due to gravity (approximately 9.8 m/s^2), h is the height, and v is the final velocity.
Since the penny's mass is not provided, we can assume it to be around 2.5 grams (or 0.0025 kg).
Now, let's plug in the values:
0.0025 kg * 9.8 m/s^2 * 406 m = (1/2) * 0.0025 kg * v^2
Simplifying the equation, we get:
3989 J = 0.00125 kg * v^2
Now, let's solve for v:
v^2 = 3989 J / 0.00125 kg
v^2 ≈ 3,191,200 m^2/s^2
Taking the square root of both sides, we find:
v ≈ 1,787 m/s
So, the speed with which the penny strikes the ground (ignoring air resistance) is approximately 1,787 m/s. That's quite a speedy penny! Just be careful it doesn't give you a penny for your thoughts too forcefully when it lands!
To find the speed with which the penny strikes the ground, we can use the equation of motion for free-falling objects. The equation is:
v^2 = u^2 + 2as
Where:
v = final velocity (the speed with which the penny strikes the ground)
u = initial velocity (which is 0 m/s since the penny is dropped from rest)
a = acceleration due to gravity (which is approximately 9.8 m/s^2)
s = distance (the height of the building, in this case)
Plugging in the given values:
v^2 = 0^2 + 2(9.8)(406)
Simplifying:
v^2 = 2(9.8)(406)
v^2 = 2(3976.8)
v^2 = 7953.6
Taking the square root of both sides:
v = sqrt(7953.6)
v ≈ 89.1 m/s
Therefore, the speed with which the penny strikes the ground is approximately 89.1 m/s.
To find the speed with which the penny strikes the ground, we can make use of the kinematic equations of motion. In this case, we'll use the following equation:
v^2 = u^2 + 2as
Where:
v is the final velocity
u is the initial velocity
a is the acceleration
s is the distance
In this scenario, the penny is dropped from rest, implying that the initial velocity (u) is 0 m/s. The distance (s) is given as the height of the building, which is 406 m. The acceleration due to gravity is approximately 9.8 m/s^2.
Substituting the known values into the equation, we get:
v^2 = 0^2 + 2 * 9.8 * 406
v^2 = 0 + 8004.4
v^2 = 8004.4
v = √8004.4
Using a calculator or an online tool to calculate the square root, we get:
v ≈ 89.49 m/s (rounded to two decimal places)
Therefore, the speed with which the penny strikes the ground is approximately 89.49 m/s.