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

v = 11.6 m/s

Step-by-step 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