A soccer player kicks a 0.43 kg ball straight up to see how far it will rise. The ball starts with a velocity of 16m/s at a height of 0.85m above the ground. It rises then falls straight down to the ground.
a) what is the total energy of the ball just after it is kicked?
b) How high above its original position does the ball rise?
c) What is the velocity of the ball when it hits the ground?
total energy= PE + KE. The question on PE, relative to what point? ground? Assuming ground
total initial energy= mh*.85 + 1/2 m 16^2
b( at the top, KE is zero...Pe = intial energy, so figure h
c. final KE= initial total energy, solve for v.
Bull
To solve these questions, we can use the principles of mechanical energy conservation.
a) The total energy of the ball just after it is kicked can be calculated as the sum of its kinetic energy and potential energy. The kinetic energy (KE) can be determined using the formula KE = (1/2)mv^2, where m is the mass of the ball and v is its velocity. The potential energy (PE) is given by PE = mgh, where h is the height above the ground.
Given that the mass of the ball is 0.43 kg, and the initial velocity is 16 m/s, we can calculate the kinetic energy as follows:
KE = (1/2)mv^2
= (1/2)(0.43 kg)(16 m/s)^2
= 55.04 J
The potential energy at a height of 0.85 m can be calculated as:
PE = mgh
= (0.43 kg)(9.8 m/s^2)(0.85 m)
= 3.96 J
Therefore, the total energy is the sum of the kinetic and potential energy:
Total energy = KE + PE
= 55.04 J + 3.96 J
= 59 J
b) The maximum height above its original position the ball rises can be calculated using the conservation of mechanical energy. At the highest point, all of the initial kinetic energy should be converted into potential energy.
So, the kinetic energy at the highest point is zero, and the potential energy is equal to the total energy.
Potential energy = Total energy
= 59 J
Using the potential energy formula PE = mgh, we can find h:
h = PE / (mg)
= 59 J / (0.43 kg × 9.8 m/s^2)
≈ 14.1 m
Therefore, the ball rises approximately 14.1 m above its original position.
c) When the ball hits the ground, its height above the ground is 0. Hence, the potential energy is zero. At this point, all of the energy is in the form of kinetic energy.
Therefore, we can equate the kinetic energy to the total energy to find the velocity of the ball just before hitting the ground:
KE = Total energy
(1/2)mv^2 = 59 J
Plug in the values:
(1/2)(0.43 kg)v^2 = 59 J
Solving for v:
v^2 = (59 J) × (2 / 0.43 kg)
≈ 546.28 m^2/s^2
v ≈ √546.28
≈ 23.38 m/s
So, the velocity of the ball when it hits the ground is approximately 23.38 m/s.
To answer these questions, we need to analyze the energy changes and motion of the ball.
a) The total energy of the ball just after it is kicked consists of its kinetic energy and its potential energy.
To calculate the kinetic energy (KE), we need the mass (m) and velocity (v) of the ball:
KE = 0.5 * m * v^2
The potential energy (PE) of the ball at 0.85m above the ground can be calculated using its height (h) and mass (m) as:
PE = m * g * h
Where g is the acceleration due to gravity (approximately 9.8 m/s^2).
To find the total energy, we sum up the kinetic energy and potential energy:
Total Energy = Kinetic Energy + Potential Energy
b) To determine how high above its original position the ball rises, we need to find its maximum height (h').
Given the initial velocity (v) and height above the ground (h), we can use the equation of motion to find the maximum height:
v^2 = u^2 + 2as
Where u is the initial velocity, a is the acceleration (which is -g due to gravity), and s is the displacement (maximum height).
c) To find the velocity of the ball when it hits the ground, we need to consider that the ball falls under gravity.
Using the equation of motion, we have:
v^2 = u^2 + 2gs
Where u is the initial velocity (which is zero when the ball reaches its maximum height), g is the acceleration due to gravity, and s is the displacement (the distance the ball falls, which is equal to the initial height).
Now let's use these equations to find the answers to your questions.