Suppose that a 112.5 kg football player running at 8.5 m/s catches a 0.44 kg ball moving at a speed of 25 m/s with his feet off the ground, while both of them are moving horizontally.
Calculate the change in kinetic energy of the system, in joules, after the player catches the ball.
Calculate the change in kinetic energy of the system, in joules, in this case.
To calculate the change in kinetic energy of the system, we need to find the initial kinetic energy (before catching the ball) and the final kinetic energy (after catching the ball) and then subtract the initial value from the final value.
Initial Kinetic Energy (before catching the ball): The initial kinetic energy of the football player can be calculated using the formula:
Initial Kinetic Energy = (1/2) * mass * velocity^2
For the football player: Mass (m1) = 112.5 kg Velocity (v1) = 8.5 m/s
Initial Kinetic Energy of the football player = (1/2) * 112.5 kg * (8.5 m/s)^2
Initial Kinetic Energy of the ball (before catching): The initial kinetic energy of the ball can be calculated using the same formula:
Initial Kinetic Energy = (1/2) * mass * velocity^2
For the ball: Mass (m2) = 0.44 kg Velocity (v2) = 25 m/s
Initial Kinetic Energy of the ball = (1/2) * 0.44 kg * (25 m/s)^2
Final Kinetic Energy (after catching the ball): After the football player catches the ball, both the player and the ball move together with the same horizontal velocity. So, we can consider them as one combined system with a total mass of:
Total Mass (m_total) = m1 + m2 Total Mass (m_total) = 112.5 kg + 0.44 kg
Now, the final velocity (v_final) of the combined system is the same as the initial velocity of the ball (25 m/s), since both are moving together horizontally after the catch.
Final Kinetic Energy = (1/2) * m_total * v_final^2
Final Kinetic Energy = (1/2) * (112.5 kg + 0.44 kg) * (25 m/s)^2
Now, we can calculate the change in kinetic energy:
Change in Kinetic Energy = Final Kinetic Energy - Initial Kinetic Energy
Change in Kinetic Energy = [(1/2) * (112.5 kg + 0.44 kg) * (25 m/s)^2] - [(1/2) * 112.5 kg * (8.5 m/s)^2]
Calculate the values:
Change in Kinetic Energy = [0.5 * 112.94 kg * 625 m^2/s^2] - [0.5 * 112.5 kg * 72.25 m^2/s^2]
Change in Kinetic Energy ≈ 35143.75 J - 4505.31 J
Change in Kinetic Energy ≈ 30638.44 Joules
So, the change in kinetic energy of the system after the player catches the ball is approximately 30,638.44 Joules.
Suppose that a 112.5 kg football player running at 8.5 m/s catches a 0.44 kg ball...
Clear Answer and steps plz, I solved it many times but never worked out (13%) Problem 6: Suppose that a 110 kg football player running at 7.5 m/s catches a 0.49 kg ball moving at a speed of 25 m/s with his feet off the ground, while both of them are moving horizontally 2500 Part (a) Calculate the final speed of the player in meters per second, if the ball and player are initially moving in the same direction. *...
1. A football player of mass 90 kg is running at 10 m/s and catches a football of mass
Kyle, a 100.0 kg football player, leaps straight up into the air (with no horizontal velocity) to catch a pass. He catches the 0.430 kg ball precisely at the peak of his jump, when he is 0.386 meters off the ground. He hits the ground 0.0295 meters away from where he leapt. If the ball was moving horizontally when it was caught, how fast was the ball traveling? ball's speed: m/s
Kyle, a 100.0 kg football player, leaps straight up into the air (with no horizontal velocity) to catch a pass. He catches the 0.430 kg ball precisely at the peak of his jump, when he is 0.488 meters off the ground. He hits the ground 0.0320 meters away from where he leapt. If the ball was moving horizontally when it was caught, how fast was the ball traveling? ball's speed:
A 70.0 kg football player leaps straight up into the air (with no horizontal velocity) to catch a pass. He catches the 0.430 kg ball precisely at the peak of his jump, when he is 0.538 meters off the ground. He hits the ground 0.0384 meters away from where he leapt. If the ball was moving horizontally when it was caught, how fast was the ball travelling?
A 70.0 kg football player leaps straight up into the air (with no horizontal velocity) to catch a pass. He catches the 0.430 kg ball precisely at the peak of his jump, when he is 23.2 inches off the ground. He hits the ground 1.11 inches away from where he leapt. If the ball was moving horizontally when it was caught, how fast was the ball travelling?
A 60.0 kg football player leaps straight up into the air (with no horizontal velocity) to catch a pass. He catches the 0.430 kg ball precisely at the peak of his jump, when he is 0.589 meters off the ground. He hits the ground 0.0371 meters away from where he leapt. If the ball was moving horizontally when it was caught, how fast was the ball travelling?
Kyle, a 95.0 kg95.0 kg football player, leaps straight up into the air (with no horizontal velocity) to catch a pass. He catches the 0.430 kg0.430 kg ball precisely at the peak of his jump, when he is 0.386 meters0.386 meters off the ground. He hits the ground 0.0371 meters0.0371 meters away from where he leapt. If the ball was moving horizontally when it was caught, how fast was the ball traveling?
5) Fred (mass 60 kg) is running with the football at a speed of 6.0 m/s when he is met head-on by Brutus (mass 120 kg), who is moving at 4.0 m/s. Brutus grabs Fred in a tight grip, and they fall to the ground. Which way do they slide and how far? The coefficient of kinetic friction between football uniforms and the ground is 0.30.
Football Collision A 96 kg running back, moving at 5.29 m/s, runs into a 109 kg defender who is initially at rest. What is the speed of the players just after their perfectly inelastic collision? Incorrect. What is the change in their total kinetic energy due to this totally inelastic collision?