A 71.1 kg man steps out a window and falls (from rest) 1.24 m to a sidewalk. What is his speed just before his feet strike the pavement?
(1/2) m v^2 = m g h
m cancels of course (Galileo)
v = sqrt (2 g h)
remember that
v = sqrt (2*9.81*1.24)
v = 4.93 m/s
To find the speed of the man just before his feet strike the pavement, we can use the principle of conservation of energy.
The potential energy of an object at height h is given by the equation P.E. = mgh, where m is the mass of the object, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height.
Here, the potential energy is converted into kinetic energy as the man falls. The kinetic energy of an object is given by the equation K.E. = (1/2)mv^2, where v is the speed of the object.
Since energy is conserved, we can equate the potential energy before falling with the kinetic energy just before hitting the ground:
mgh = (1/2)mv^2
We can cancel out the mass, m, from both sides of the equation:
gh = (1/2)v^2
Now, we can solve for v by rearranging the equation:
v^2 = 2gh
Taking the square root of both sides:
v = sqrt(2gh)
Substituting the given values:
v = sqrt(2 * 9.8 m/s^2 * 1.24 m)
v = sqrt(24.264 m^2/s^2)
v = 4.93 m/s
Therefore, the man's speed just before his feet strike the pavement is approximately 4.93 m/s.