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 19.2 inches off the ground. He hits the ground 1.61 inches away from where he leapt. If the ball was moving horizontally when it was caught, how fast was the ball travelling?
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To find the speed at which the ball was traveling horizontally when it was caught, we can use the principle of conservation of momentum.
First, let's calculate the total initial momentum of the system. The football player's momentum can be calculated by multiplying his mass (70.0 kg) by his vertical velocity when he catches the ball. Since he reached the peak of his jump, his vertical velocity at that point is zero.
So the football player's initial momentum is 0 kg*m/s.
Next, let's calculate the momentum of the ball when it was caught. The momentum of the ball can be calculated by multiplying its mass (0.430 kg) by its horizontal velocity (which we want to find).
So the ball's momentum when it was caught is 0.430 kg * v, where v is the horizontal velocity.
According to the principle of conservation of momentum, the total momentum before and after the catch should be equal. Therefore, we can set up the equation:
0 kg*m/s + 0.430 kg * v = 0 kg*m/s.
Simplifying the equation, we get:
0.430 kg * v = 0 kg*m/s.
Since the mass of the ball is not zero, the only way for this equation to hold is if the horizontal velocity (v) is zero.
Therefore, we can conclude that the ball was not moving horizontally when it was caught, and its horizontal velocity was zero.
Please note that the distance the football player lands away from where he leapt and the vertical distance he jumps do not affect the horizontal velocity of the ball, which remains zero.