A skateboarder rolls off a horizontal ledge that is 1.32m high and lands 1.88m from the base of the ledge. What was the initial velocity? (Unit=m/s)

Velocity = 3.622 m/s

Explanation:

Given:

Starting height = 1.32 m

Distance = 1.88 m

Find:

Velocity

Computation:

Using 2nd equation of motion.

s = ut + [1/2]gt²  , where g = 9.8 m/s² and u = 0

1.32 = [1/2][9.8]t²

Time (t) = 0.519 sec

We know that

Velocity = Distance / Time

Velocity = 1.88 / 0.519

Velocity = 3.622 m/s

the diagram which shows forces as either balanced = same speed or unbalanced = accelerarting

in simple terms, the principle states that the buoyancy force on an object is equal to the weight of the fluid displaced by the object, or the density of the fluid multiplied by the submerged volume times the gravitational acceleration, g. this is also known as upthrust.