Kyle, a 85.0 kg 85.0 kg football player, leaps straight up into the air (with no horizontal velocity) to catch a pass. He catches the 0.430 kg 0.430 kg ball precisely at the peak of his jump, when he is 0.437 meters 0.437 meters off the ground. He hits the ground 0.0295 meters 0.0295 meters away from where he leapt. If the ball was moving horizontally when it was caught, how fast was the ball traveling?

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JemdetNasr JemdetNasr · Answer 1 · 2019-12-21T06:57:19+01:00

Answer:

[tex]19.6 ms^{-1}[/tex]

Explanation:

Consider the motion of the player+ball together after catch along the vertical direction.

[tex]V_{oy}[/tex] = speed of player+ball together after catch = 0 m/s

[tex]a_{y}[/tex] = acceleration due to gravity = [tex]9.8 m/s^{-2}[/tex]

[tex]t[/tex] = time of travel

[tex]y[/tex] = vertical displacement = - 0.437 m

Using the kinematics equation that suits the above list of data, we have

[tex]y = V_{oy} t + (0.5) a_{y} t^{2} \\- 0.437 = (0) t + (0.5) (- 9.8) t^{2}\\t = 0.2986 s[/tex]

Consider the motion of the player+ball together after catch along the horizontal direction.

[tex]V[/tex] = speed of player+ball together after catch

[tex]X[/tex] = Horizontal distance traveled = 0.0295 m

[tex]t[/tex] = time taken = 0.0295 m

Since there is no acceleration along the horizontal direction, we have

[tex]X = V t \\0.0295 = V (0.2986)\\V = 0.0988 ms^{-1}[/tex]

[tex]v[/tex] = speed of the ball before catch

[tex]M[/tex] = mass of the player = 85 kg

[tex]m[/tex] = mass of the ball = 0.430 kg

Using conservation of momentum

[tex]m v = (m + M) V\\(0.430) v = (0.430 + 85)(0.0988)\\v = 19.6 ms^{-1}[/tex]