Question

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 0.437 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 traveling?

Answer 1

The ball was traveling with a vertical velocity of approximately 3.02 m/s.

Step-by-step explanation:

To calculate the velocity of the ball, we can apply the principle of conservation of momentum. Since there is no horizontal velocity, the total horizontal momentum before and after the catch is zero. Therefore, the horizontal velocity of the ball after it is caught is also zero.

To calculate the vertical velocity of the ball, we can use the equation of motion:

Vf^2 = Vi^2 + 2ad

where Vf is the final velocity, Vi is the initial velocity, a is the acceleration, and d is the distance travelled vertically. Plugging in the given values, we have:

0 = Vi^2 + 2*(-9.8)*0.437

Solving for Vi, we get:

Vi = sqrt(2*9.8*0.437)

Vi ≈ 3.02 m/s

Therefore, the ball was traveling with a vertical velocity of approximately 3.02 m/s.