Question

A 100 kg skateboarder sits at the top of a 10 m halfpipe, ready to go down the ramp. What will be his velocity when

he is 4 m off of the ground?

Answer 1

v = 10.85 m/s

Step-by-step explanation:

We will apply the law of conservation of energy to the skateboarder. Neglecting the frictional effects, the law of conservation of energy can be written as:

`Loss\ in\ Potential\ Energy\ of\ Skateboarder = Gain\ in\ Kinetic\ Energy\ of\ Skateboarder`
`mg\Delta h = (1)/(2)mv^2\\\\v^2 = 2g\Delta h\\v = √(2g\Delta h) \\`

where,

v = velocity of skateboarder = ?

g = acceleration due to gravity = 9.81 m/s²

Δh = change in height = 10 m - 4 m = 6 m

Therefore,

`v = \sqrt{{(2)(9.81\ m/s^2})({6\ m})}}`

v = 10.85 m/s