Question

The log left the cliff with a horizontal velocity of 7.5 m/s. When it reached the water, it had a vertical velocity of -8.9 m/s. What was the log's speed (two dimensional) before it entered the water

Answer 1

Answer:

v = 11.6 m/s

Explanation:

• Assuming no other forces acting upon the log than gravity (which acts only in the vertical direction)  in the horizontal direction, velocity must be constant at any point.
• So, before it entered the water, its velocity has two components, as follows:
• vₓ = 7.5 m/s
• vy = -8.9 m/s
• We can find the magnitude of the velocity vector, just applying Pythagorean Theorem to both components, as follows:

       $v = \sqrt{v_{x}^{2} + v_{y}^{2}} = \sqrt{((7.5m/s)^{2} + (-8.9m/s)^{2}$

       ⇒ v = 11.6 m/s