Question

A skateboarder rolls horizontally off the top of a staircase at a speed of 18 ms-1 and at the bottom of the stairs which has a horizontal length of 9 m as
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Calculate the skater's vertical displacement during the jump

Answer 1

Motion of a skateboarder is a two dimensional motion experienced by an object or particle that is subjected near the Earth's surface and moves along a curved path under the influence of gravity only. 

 Given that in motion of the skateboarder the horizontal component of the velocity remain constant throughout his motion, there is no acceleration along horizontal path.

Skateboarder jumps horizontally. So initial velocity has only horizontal component $(v_{0y}=0)$

The equations of motion

$$x=v_{0x}t$$ $$y={gt^2\over 2}$$

Then

$$t={x\over v_{ox}}$$

$$y={gx^2\over 2v_{0x}^2}$$

$x=9m ,v_{0x}=18\ ms^{-1},g=9.81\ m/s^{-2}$

$$y={9.81\ m/s^{-2}(9m)^2\over 2(18\ ms^{-1})^2}\approx1.226\ m$$

 The skater's vertical displacement during the jump is 1.226 m down.