# The height of a slope making an angle of 30 degrees with the horizontal is 36 m. What is the velocity of an object that slides down from the top.

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You need to apply the principle of conservation of mechanical energy such that: `mgh = (mv^2)/2 =gt 2gh = v^2 =gt v = +-sqrt(2gh).`

Considering the motion of object frictionless, hence `v = sqrt(2gh).`

The problems provides the value of height, hence the angle made by slope to x axis is of no need in this problem. You need to know the angle only if the length of slope is known but the height is not given.

Substituting is`9.8 m/(s^2)` for g and 36 m for h yields:v = `sqrt(2*9.8*36) ` => `v~~ 26.56 m/s`

**Hence, evaluating the velocity of object sliding down from the top of slop yields: v~~ 26.56 m/s.**

Height of incline from Horizantal = 36 meters

Angle of incline with horizontal = 30 degrees

Length of incline = Height / sin(Slope) = 36 / Sin(30) = 72 meters

Gravitational pull in vertical direction = 9.8 m/s2

Acceleration in the direction of incline = 9.8 * sin(30) = 4.9 m/s2

using formula v^2-u^2 = 2*a*s

where v = final velocity u = initial velocity = 0

a = acceleration = 4.9 m/s2

and s = length traversed = 72 meters

we get: v^2 - 0 = 2*4.9*72 = 705.6

or v = sqrt(705.6) = 26.56

Thus the **velocity of the object = 26.56 m/s**

*This velocity is not in the vertical direction but in the direction of the slope as the direction of the acceleration was in the same direction.*

The above also explains your question made to sciencesolve as to why the velocity is not vertical.

Why is the direction of the velocity not vertical, though only a vertical force acts on it? Could you please explain that, thank you.