A penny is dropped from a 50 m high bridge, how fast was the penny moving when it hit the water?
V^2 = Vo^2 + 2g*h
V^2 = 0 + 19.6*50. Solve for V.
To determine the speed of the penny when it hits the water, we can use the laws of physics and apply the principle of conservation of energy. The initial potential energy from being at a height of 50 m will be converted entirely into kinetic energy just before the penny hits the water.
To calculate the speed, we need to know the acceleration due to gravity (g), which is approximately 9.8 m/s^2. We can use the following equation:
Potential energy (PE) = Kinetic energy (KE)
PE = mgh
KE = (1/2)mv^2
Where:
m = mass of the penny (which we can assume to be constant)
g = acceleration due to gravity
h = height of the bridge (50 m in our case)
v = velocity (speed) of the penny when it hits the water
Setting the potential energy equal to the kinetic energy:
mgh = (1/2)mv^2
Since the mass and the mass cancel out on both sides:
gh = (1/2)v^2
Now, isolate v:
v^2 = 2gh
Taking the square root of both sides:
v = √(2gh)
Substituting the known values:
v = √(2 * 9.8 m/s^2 * 50 m)
v ≈ 31.3 m/s
Therefore, when the penny hits the water, it is moving at a speed of approximately 31.3 m/s.