When a skateboarder ollies, you can think of it as moving along horizontally and then jumping straight up.
If a skateboarder is rolling along at 10m/s and then ollies 80cm up into the air.What is his velocity when he leaves the ground and what horizontal distance does he cover before landing?
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When a skateboarder ollies, it can be considered as moving along horizontally and then jumping straight up.
The skateboarder ollies up a distance 80 cm = 0.8 m into the air. Let the initial vertical velocity when the skateboarder jumps be V. The velocity decreases at a rate equal to 9.8 m/s^2 and at the highest point it is equal to 0. This gives: 0 = V - 9.8*t
=> t = V/9.8
The highest point is at a height of 0.8 m, 0 - V^2 = -2*9.8*0.8
=> V^2 = 15.68
=> V = 3.96 m/s
The velocity when the skateboarder leaves the ground is 3.96 m/s
t = V/9.8 = 0.4 s
An equal duration of time is spent in going up as is spent in coming down.
The horizontal distance traveled while the skateboarder is in the air is the distance traveled in 0.8 s and is equal to 8.08 m.
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