A skateboarder rolls horizontally off the top of a staircase and lands at the bottom. The staircase has a horizontal length of 12.0\,\text m12.0m12, point, 0, start text, m, end text with a vertical drop of 2.00\,\text m2.00m2, point, 00, start text, m, end text. We can ignore air resistance.

Explanation:

Complete question:

A skateboarder rolls horizontally off the top of a staircase and lands at the bottom. The staircase has a horizontal length of 12.0m with a vertical drop of 2.0m. Calculate the speed of the skateboard. We can ignore air resistance.

Using the following equations of motion;

Horizontal displacement

x = vt 1

Vertical component;

y = 1/2gt² 2

v is the speed

t is the time

g is the acceleration due to gravity

Frim 1;

t = x/v 3

Substitute equation 3 into 2;

From 2; y =  1/2gt² .

y = 1/2 g(x/v)²

y = 1/2 gx²/v²

y =  gx²/2v²

v² = gx²/2y

v = √gx²/2y

Given;

x = 12m

y = 2m

g = 9.8m/s²

Substitute the given parameters into the give equation;

v = √(9.8)(12)²/2(4)

v =  √(9.8)(144)/8

v =  √1411.2/8

v = √176.4

v = 13.28m/s

Hence the approximate speed of the skateboard is 13.28m/s

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