quarterback. The mass of the ball is 0.41 kg it has a speed of 22.0 m/s just before the player caught it. Assuming that the football player catches the ball with his feet off the ground and both the player and ball are moving horizontally, calculate the final velocity of the ball and player

Question: A football player of mass 107 kg, is running at a speed of 6.7 m/s when it catches a ball thrown by the quarterback. The mass of the ball is 0.41 kg it has a speed of 22.0 m/s just before the player caught it. Assuming that the football player catches the ball with his feet off the ground and both the player and ball are moving horizontally, calculate the final velocity of the ball and player

6.76 m/s

Explanation:

From the question above,

Applying the law of conservation of momentum

Total momentum before collision = Total momentum after collisiom

Note: This type of collision is an elastic collision, and so the player and the ball moves with a common velocity after collision.

Mv+mu = V(M+m) Equation 1

Where M = mass of the player, v = initial velocity of the player, m = mass of the ball, u = initial velocity of the ball, V = Final velocity of the ball and the player

make V the subject of the equation

V = (Mv+mu)/(M+m) Equation 2

Given: M = 107 kg, m = 0.41 kg, v = 6.7 m/s, u = 22 m/s

Substitute into equation 2

V =[(107×6.7)+(0.41×22)]/(107+0.41)

V =  (716.9+9.02)/107.41

V = 6.76 m/s.

im guessing the west of station b

a) the event horizon, singularity, and the chute located between the two.