Liz asked in Science & MathematicsPhysics · 9 months ago

The drawing shows a person who, starting from rest at the top of a cliff, swings down at the end of a rope, releases it...?

The drawing shows a person who, starting from rest at the top of a cliff, swings down at the end of a rope, releases it, and falls into the water below. There are two paths by which the person can enter the water. Suppose he enters the water at a speed of 18.5 m/s via path 1. How fast is he moving on path 2 when he releases the rope at a height of 3.14 m above the water? Ignore the effects of air resistance.

1 Answer

Relevance
  • NCS
    Lv 7
    9 months ago
    Favorite Answer

    The drawing might be important, but without it I'll wager that he enters the water at 18.5 m/s regardless of the path he takes (by conservation of energy).

    If you're asking about his speed as he releases the rope, then I guess we'd do this:

    height above water h = v² / 2g = (18.5m/s)² / 19.6m/s² = 17.5 m

    and so at a height of 3.14 m, we'd have

    V = √2gh) = √(2 * 9.8m/s² * (17.5 - 3.14)m) = 16.7 m/s ◄

    If you find this helpful, please award Best Answer. You get points too!

Still have questions? Get your answers by asking now.