After performing a trick above the rim of a skateboard ramp, a 56 kg skateboarder lands on the ramp 3.5 m above ground level with a downward velocity of 4.0 m/s. Ep = 0.0 J Ground level. Friction...
The kinetic energy of an object with mass m, moving at v m/s is given by the formula ME = `(1/2)*m*v^2` .
This gives the increase in velocity as `sqrt((1020.8*2)/56) ~~ 6.038` m/s
The skateboarder had an initial downward speed of approximately 6 m/s when he landed on the ramp. The speed of the skateboarder at the bottom of the ramp is 10.038 m/s.