An athlete throws a basketball through a hoop. The ball leaves his hands at a height of 2.5 m with a speed of 9.0 m/s. Calculate the speed of the ball when it “swishes” through the hoop. The hoop is at a height of 3.0 m and it is 5.0 m along the court from the basketball player.
It is pretty easy...
consider energy.
The ball has an initial KE, and initial PE. The ball has a final PE you know, so calculate the final KE.
Intialtotalenergy= final PE+finalKE
I tried the method listed above but it is different from the answer given in the textbook. I hope the textbook is wrong!
To solve this problem, we can use the principle of conservation of energy. Initially, the ball is at a height of 2.5 m with a speed of 9.0 m/s, and when it goes through the hoop, it reaches a height of 3.0 m. We can assume there is no air resistance.
Step 1: Calculate the potential energy at the initial position:
Potential energy (PE) = m * g * h
Where m is the mass of the basketball, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height from the initial position to the ground.
Given that the height is 2.5 m, the mass of the basketball is not provided, so we can assume a standard basketball mass of 0.625 kg.
PE = 0.625 kg * 9.8 m/s^2 * 2.5 m = 15.31 J
Step 2: Calculate the kinetic energy at the initial position:
Kinetic energy (KE) = 1/2 * m * v^2
Where v is the initial velocity of the basketball.
Given that the velocity is 9.0 m/s, and the mass is 0.625 kg, we can calculate the kinetic energy.
KE = 1/2 * 0.625 kg * (9.0 m/s)^2 = 30.94 J
Step 3: Calculate the total mechanical energy at the initial position:
Total mechanical energy (E) = PE + KE = 15.31 J + 30.94 J = 46.25 J
Step 4: Calculate the potential energy at the hoop position:
Potential energy (PE) = m * g * h
Where h is the height from the hoop position to the ground. Given that the height is 3.0 m and the mass of the basketball is 0.625 kg, we can calculate the potential energy.
PE = 0.625 kg * 9.8 m/s^2 * 3.0 m = 18.38 J
Step 5: Calculate the kinetic energy at the hoop position:
As the ball goes through the hoop, the height decreases, and therefore the potential energy decreases. The total mechanical energy remains constant at 46.25 J (conservation of energy).
Therefore, the kinetic energy at the hoop position (KE_hoop) can be calculated as:
KE_hoop = Total mechanical energy - PE_hoop
= 46.25 J - 18.38 J
= 27.87 J
Step 6: Calculate the speed of the ball when it "swishes" through the hoop:
To calculate the speed, we can use the equation:
KE_hoop = 1/2 * m * v_hoop^2
Rearranging the equation, we can solve for v_hoop:
v_hoop = √(2 * KE_hoop / m)
Where m is the mass of the basketball.
Given that the mass is 0.625 kg and KE_hoop is 27.87 J, we can calculate the speed of the ball at the hoop position.
v_hoop = √(2 * 27.87 J / 0.625 kg) = 11.88 m/s
Therefore, the speed of the ball when it "swishes" through the hoop is 11.88 m/s.
To calculate the speed of the ball when it "swishes" through the hoop, we need to consider the conservation of energy and the motion of the ball.
First, let's find the initial kinetic energy (KE) and potential energy (PE) of the ball when it leaves the athlete's hands.
The initial potential energy is given by PE = mgh, where m is the mass of the ball, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the initial height.
PE = mgh
PE = (mass) * (9.8 m/s^2) * (2.5 m)
Next, let's find the final potential energy when the ball is at a height of 3.0 m, which is the height of the hoop.
PE_final = mgh
PE_final = (mass) * (9.8 m/s^2) * (3.0 m)
Since the ball "swishes" through the hoop without bouncing, the final kinetic energy is equal to the initial kinetic energy.
KE_initial = KE_final
The initial kinetic energy can be calculated using the equation KE = (1/2)mv^2, where m is the mass of the ball and v is the initial velocity.
KE_initial = (1/2)mv_initial^2
KE_initial = (1/2)*(mass)*(9.0 m/s)^2
Now, we can equate the initial and final kinetic energy to find the velocity (speed) of the ball as it swishes through the hoop.
(1/2)*(mass)*(9.0 m/s)^2 = (1/2)*(mass)*v_final^2
Canceling out the mass from both sides of the equation, we get:
(9.0 m/s)^2 = v_final^2
Taking the square root of both sides, we find:
v_final = 9.0 m/s
Therefore, the speed of the ball when it "swishes" through the hoop is 9.0 m/s.