During a free kick at a soccer playoff game a 0.5 kg ball at the peak of the kick 14.7 m above the field had a kinetic energy of 34.0 J. How fast was the ball booted by the player on the field?
To determine the speed of the ball booted by the player, we can use the principle of conservation of energy. At the peak of the kick, the ball will have maximum potential energy and zero kinetic energy. And when it reaches the field, it will have maximum kinetic energy and zero potential energy.
We can use the formula for kinetic energy:
KE = 0.5 * m * v^2
Where:
KE is the kinetic energy
m is the mass of the ball
v is the velocity (speed) of the ball
Given:
m = 0.5 kg
KE = 34.0 J
Rearranging the formula, we can solve for v:
v^2 = (2 * KE) / m
Substituting the given values:
v^2 = (2 * 34.0 J) / 0.5 kg
v^2 = 68.0 J / 0.5 kg
v^2 = 136 m^2/s^2
Taking the square root of both sides to solve for v:
v = √136 m/s
v ≈ 11.66 m/s
Therefore, the ball was booted by the player at a speed of approximately 11.66 m/s.
To determine the velocity at which the ball was kicked, we can use the equation for kinetic energy:
Kinetic Energy (KE) = (1/2) * mass * velocity^2
Given:
Mass (m) = 0.5 kg
Height (h) = 14.7 m
Kinetic Energy (KE) = 34.0 J
First, we need to find the potential energy of the ball when it was at its peak. The potential energy is given by the equation:
Potential Energy (PE) = mass * gravity * height
Since the ball is at its peak, its potential energy is equal to its kinetic energy:
PE = KE
mass * gravity * height = KE
Substituting the given values:
0.5 kg * 9.8 m/s^2 * 14.7 m = 34.0 J
Now, solve for the velocity (v):
v = √((2 * KE) / mass)
Substituting the given value for KE and mass:
v = √((2 * 34.0 J) / 0.5 kg)
v = √(68 J / 0.5 kg)
v = √(136 m^2/s^2 / 0.5 kg)
v = √(272 m^2/s^2 / kg)
v = √(272) m/s
v ≈ 16.49 m/s
Therefore, the ball was kicked at a speed of approximately 16.49 m/s by the player on the field.