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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.
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