During a free kick at a soccer playoff game a 0.5 kg ball at the peak of the kick 16.6 m above the field had a kinetic energy of 35.4 J. How fast was the ball booted by the player on the field?
To find the speed at which the ball was kicked by the player, we need to use the principle of conservation of energy.
The kinetic energy (KE) of the ball at the peak of the kick is given as 35.4 J.
The potential energy (PE) of the ball at the peak of the kick can be calculated using the formula:
PE = mgh
Where m is the mass of the ball (0.5 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the height (16.6 m).
PE = (0.5 kg) * (9.8 m/s^2) * (16.6 m)
PE = 81.74 J
According to the principle of conservation of energy, the total mechanical energy of the system remains constant. Therefore, the sum of the kinetic energy and potential energy at any point in time should be equal to the total mechanical energy of the system.
So, the total mechanical energy (E) of the system can be calculated as:
E = KE + PE
E = 35.4 J + 81.74 J
E = 117.14 J
Now, at the peak of the kick, the kinetic energy is maximum and the potential energy is zero. Therefore, the sum of kinetic energy and potential energy is equal to the total mechanical energy.
35.4 J + 81.74 J = 117.14 J
So, the speed at which the ball was kicked by the player can be calculated using the formula for kinetic energy:
KE = 0.5mv^2
Where m is the mass of the ball (0.5 kg) and v is the velocity (speed) of the ball.
35.4 J = 0.5 * (0.5 kg) * v^2
70.8 J = 0.25 kg * v^2
v^2 = (70.8 J) / (0.25 kg)
v^2 = 283.2 m^2/s^2
v = √(283.2 m^2/s^2)
v ≈ 16.8 m/s
Therefore, the ball was kicked at a speed of approximately 16.8 m/s by the player on the field.
V^2 = Vo^2 + 2g*h = 0 at peak.
Vo^2 - 19.6*16.6 = 0
Vo^2 = 325.4
Vo = 18.04 m/s = Speed at which ball was kicked.