A 67.5-kg basketball player jumps vertically and leaves the floor with a velocity of 1.66 m/s upward. (Take upward as positive and forward as positive.)
(a) What impulse does the player experience?
magnitude
direction (up,down, back or zero)
(b) What force does the floor exert on the player before the jump?
magnitude
direction
(c) What is the total average force exerted by the floor on the player if the player is in contact with the floor for 0.450 s during the jump?
magnitude
direction
To find the answers to these questions, we can use the principle of impulse-momentum. Impulse is defined as the change in momentum, and it can be calculated by multiplying the force applied to an object by the time interval over which the force is applied.
(a) To find the impulse experienced by the player, we can use the equation:
Impulse = Change in momentum = m * Δv
where m represents the mass of the player and Δv represents the change in velocity. In this case, the player jumps vertically and leaves the floor with a velocity of 1.66 m/s upward. Since the player is moving upward, we need to consider the direction as positive. The initial velocity of the player before the jump is zero, so the change in velocity is 1.66 m/s. The mass of the player is given as 67.5 kg.
Impulse = 67.5 kg * 1.66 m/s = 111.21 kg·m/s
The magnitude of the impulse experienced by the player is 111.21 kg·m/s.
The direction of the impulse is upward (positive direction) since the player is moving vertically upward.
(b) To find the force exerted by the floor on the player before the jump, we can use the equation:
Impulse = Force * Time
Since the player is in contact with the floor before the jump, the impulse experienced by the player is equal to the force applied by the floor multiplied by the time of contact. The time of contact is not provided in the question, so we cannot find the exact force. However, we can say that the force exerted by the floor is equal in magnitude and opposite in direction to the impulse experienced by the player.
Therefore, the magnitude of the force exerted by the floor before the jump is 111.21 N (equal to the magnitude of the impulse experienced by the player).
The direction of the force is downward, opposite to the direction of the player's initial velocity.
(c) To find the total average force exerted by the floor on the player during the 0.450 s of contact, we can use the equation:
Average Force = Impulse / Time
The impulse experienced by the player is the same as in part (a), which is 111.21 kg·m/s. The time of contact is given as 0.450 s.
Average Force = 111.21 kg·m/s / 0.450 s ≈ 247.13 N
The magnitude of the total average force exerted by the floor on the player during the jump is approximately 247.13 N.
The direction of the force is downward, opposite to the direction of the player's initial velocity.
a)112.05 upward
b)661.5 upward