A beer is dropped from a window 30 m above the ground. How fast will it be moving just before it lands? (Neglect air resistance).
(1/2) m v^2 = m g h
so
v = sqrt (2 g h)
g = 9.81 m/s^2
h = 30 m
To determine the speed at which the beer will be moving just before it lands, we can use the principle of conservation of energy. The potential energy of the beer when it is at the window height is converted into kinetic energy just before it lands.
The potential energy (PE) is given by the equation:
PE = m * g * h
Where:
m is the mass of the beer (assumed to be known)
g is the acceleration due to gravity (approximately 9.8 m/s^2)
h is the height from which the beer is dropped (30 m)
The kinetic energy (KE) of the beer just before it lands is given by the equation:
KE = (1/2) * m * v^2
Where:
m is the mass of the beer (assumed to be known)
v is the velocity of the beer just before it lands (what we want to find)
Since the principle of conservation of energy states that energy cannot be created or destroyed, the potential energy equals the kinetic energy:
PE = KE
m * g * h = (1/2) * m * v^2
Simplifying the equation, we can solve for v:
v^2 = 2 * g * h
Taking the square root of both sides:
v = √(2 * g * h)
Plugging in the values for g (9.8 m/s^2) and h (30 m):
v = √(2 * 9.8 * 30)
Calculating this, the speed at which the beer will be moving just before it lands, neglecting air resistance, is approximately 24.2 m/s.