A ball is dropped from a height of 20 m. You can use the relation between the work done by gravity and the change in the ball's kinetic energy to determine the speed of the ball at any point in its free fall. The speed of the ball when it hits the ground =
(1/2) m v^2 = m g h
so
v = sqrt (2 g h)
if I call g = 9.81
v = 19.8 m/s
To find the speed of the ball when it hits the ground, we can use the principle of conservation of energy.
The total mechanical energy of the ball is conserved throughout its free fall, as there are no external forces doing work on the ball. Therefore, the initial potential energy of the ball at a height of 20 m is completely converted into kinetic energy as it falls.
The potential energy at the starting point is given by the formula:
Potential Energy = mass × gravitational acceleration × height
P.E. = mgh
Kinetic energy is given by the formula:
Kinetic Energy = (1/2) × mass × velocity^2
K.E. = (1/2)mv^2
According to the principle of conservation of energy:
Initial Potential Energy = Final Kinetic Energy
Therefore, we can set up the equation:
mgh = (1/2)mv^2
Mass (m) appears on both sides of the equation and cancels out. The formula simplifies to:
gh = (1/2)v^2
Now, we can rearrange the formula to solve for the final velocity (v):
v^2 = 2gh
Taking the square root of both sides, we get:
v = √(2gh)
Plugging in the values given:
g = acceleration due to gravity ≈ 9.8 m/s^2
h = height = 20 m
v = √(2 × 9.8 m/s^2 × 20 m)
v = √(392 m^2/s^2)
v ≈ 19.8 m/s
Therefore, the speed of the ball when it hits the ground is approximately 19.8 m/s.