A 92.0 kg football player at 6.50 m/s North collides with an 85.0 kg football player running at 6.00 m/s south. The 92.0 kg football player continues moving at a velocity of 2.00 m/s after the collision. What is the velocity of the 85.0 kg football player after the collision?

-1.13 m/s

Step-by-step explanation:

Parameters given:

Mass of first footballer, M = 92 kg

Initial velocity of first footballer, U = 6.5 m/s (taking North to be the +ve y axis)

Mass of second footballer, m = 85 kg

Initial velocity of second footballer, u = -6.0 m/s (taking South to be the -ve y axis)

Final velocity of first footballer, V = 2.0 m/s

We need to find the final velocity of the second football (v)

Applying the principle of conservation of momentum, we have that, in a system:

Total Initial Momentum = Total Final Momentum

(M * U) + (m * u) = (M * V) + (m * v)

(92 * 6.5) + (85 * -6) = (92 * 2) + (85 * v)

598 - 510 = 184 + 85v

88 = 184 + 85v

88 - 184 = 85v

-96 = 85v

=> v = -96 / 85

v = -1.13 m/s

The final velocity of the 85 kg player is -1.13 m/s i.e 1.13 m/s South

your answer is the frist one

hop i

answer:   28.56 feet (approx)

step-by-step explanation:

since, triangle abc has measure of angle c equal to 55 degrees, measure of angle abc equal to 90 degrees, and length of bc equal to 20 feet.

we have to find out the measurement of segment ab.

therefore, tan 55° =

but bc = 20 feet.

⇒ tan 55 ° =  

⇒ ab = 20 × tan 55 °

⇒ ab = 20 × 1.42814800674

⇒ ab = 28.5629601348   ≈ 28.56 feet