Question

A climber throws a rope with a velocity of 6.6 m/s downward to another climber 8.3 m directly below. What is the magnitude of the rope's velocity when it reaches the lower climber? Assume no air resistance.

Answer 1

When the climber throws the rope downward with a velocity of 6.6 m/s to another climber 8.3 m directly below, the magnitude of the rope's velocity when it reaches the lower climber is approximately 14.36 m/s.

Step-by-step explanation:

To solve this problem, we can use the principles of kinematics. We can consider the downward direction as positive, since the rope is thrown downwards. The initial velocity of the rope is 6.6 m/s in the downward direction.

Using the equation: v² = u² + 2as (where v is final velocity, u is initial velocity, a is acceleration, and s is displacement), we can calculate the final velocity:

v² = (6.6 m/s)² + 2(-9.8 m/s²)(-8.3 m)

v² = 43.56 m²/s² + 162.68 m²/s²

v² = 206.24 m²/s²

v = √(206.24 m²/s²) ≈ 14.36 m/s

Therefore, the magnitude of the rope's velocity when it reaches the lower climber is approximately 14.36 m/s.