A 47.0 g golf ball is driven from the tee with an initial speed of 54.0 m/s and rises to a height of 23.2 m.
(a) Neglect air resistance and determine the kinetic energy of the ball at its highest point.
What is its speed when it is 9.0 m below its highest point
To determine the kinetic energy of the golf ball at its highest point, we can use the principle of conservation of energy. At its highest point, all of its initial kinetic energy will be converted into potential energy.
The formula for kinetic energy (KE) is given by:
KE = 1/2 * mass * velocity^2
Given:
Mass of the golf ball (m) = 47.0 g = 0.0470 kg
Initial speed (v) = 54.0 m/s
Step 1: Calculate the initial kinetic energy (KE_initial) using the formula:
KE_initial = 1/2 * m * v^2
KE_initial = 1/2 * 0.0470 kg * (54.0 m/s)^2
KE_initial ≈ 69.0984 J (to four significant figures)
Therefore, the initial kinetic energy of the ball is approximately 69.0984 Joules.
Step 2: Determine the speed of the ball when it is 9.0 m below its highest point.
Since we are neglecting air resistance, we can use the principle of conservation of energy to determine the speed at a specific height.
At any height below its highest point, the total mechanical energy (sum of kinetic and potential energy) is constant.
So, we can equate the initial kinetic energy to the sum of the kinetic and potential energy at any given height.
KE_initial = KE_final + PE
Where KE_final is the final kinetic energy at the given height and PE is the potential energy at that height.
Given:
Height (h) = 9.0 m
Step 3: Calculate the potential energy (PE) at a height of 9.0 m using the formula:
PE = mass * g * h
Where g is the acceleration due to gravity (9.8 m/s^2).
PE = 0.0470 kg * 9.8 m/s^2 * 9.0 m
PE = 4.1076 J (to four significant figures)
Step 4: Rearrange the equation from Step 2 to solve for the final kinetic energy (KE_final):
KE_final = KE_initial - PE
KE_final = 69.0984 J - 4.1076 J
KE_final ≈ 64.9908 J (to four significant figures)
Step 5: Finally, we can find the speed (v_final) when it is 9.0 m below its highest point using the formula for kinetic energy:
KE_final = 1/2 * m * v_final^2
Rearranging the equation, we get:
v_final = sqrt(2 * KE_final / m)
v_final = sqrt(2 * 64.9908 J / 0.0470 kg)
v_final ≈ 76.1239 m/s (to four significant figures)
Therefore, the speed of the golf ball when it is 9.0 m below its highest point is approximately 76.1239 m/s.
To determine the kinetic energy of the golf ball at its highest point, we can use the principle of conservation of mechanical energy. The mechanical energy of the ball, which consists of its kinetic energy (KE) and potential energy (PE), remains constant throughout its motion.
At the highest point, the ball's potential energy is at its maximum, and its kinetic energy is zero (since it momentarily stops). Therefore, we can calculate the potential energy at the highest point and equate it to the kinetic energy at that point.
Potential energy (PE) = mgh
where m is the mass of the golf ball (47.0 g = 0.0470 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the height (23.2 m).
PE = (0.0470 kg)(9.8 m/s^2)(23.2 m) = 10.9792 J
Since the mechanical energy is conserved, the kinetic energy at the highest point is equal to the potential energy:
KE = 10.9792 J
Now, to determine the speed of the ball when it is 9.0 m below its highest point, we need to consider the conservation of mechanical energy again. At this position, the ball has some potential energy and some kinetic energy.
Let's assume that the speed when it is 9.0 m below the highest point is v.
Potential energy (PE) = mgh
PE = (0.0470 kg)(9.8 m/s^2)(9.0 m) = 4.0128 J
Since the mechanical energy is conserved, the sum of the potential and kinetic energy at this point is equal to the initial potential energy:
PE + KE = 10.9792 J
KE + 4.0128 J = 10.9792 J
KE = 10.9792 J - 4.0128 J = 6.9664 J
The kinetic energy at 9.0 m below the highest point is 6.9664 J.
To find the speed (v), we can use the kinetic energy equation:
KE = (1/2)mv^2
6.9664 J = (1/2)(0.0470 kg)(v^2)
v^2 = (6.9664 J * 2) / 0.0470 kg
v^2 = 296 J / 0.0470 kg
v^2 = 6297.87 m^2/s^2
v ≈ √6297.87 m^2/s^2
v ≈ 79.38 m/s (rounded to two decimal places)
Therefore, the speed of the golf ball when it is 9.0 m below its highest point is approximately 79.38 m/s.
43.17
You know the initial KEnergy, and the initial PEnergy. You know the PEnergy at the highest point, and also the PEnergy nine meters below that.
Conservation of energy:
InitialPE+InitialKE=finalPE+finalKE
Use that to find the answers.