Acceleration of gravity
Physics

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An object is released from rest and falls in free fall motion. The speed v of the object after it has fallen a distance y is given by
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Given v^2=2gy ....(1)
Differentiating this above equation with respect to v, we then have
2v=2g(dy/dv)+2y(dg/dv)
2vdv=2g(dy)+2y(dg)
Dividing both sides with v^2, we have
2(dv/v)=(2g/v^2)dy+(2y/v^2)dg
Grom equation (1), we have
2(dv/v)=(dy/y)+(dg/g) (2)
Given that dy/y=0.05 and dv/v=0.0413
Substituting this in equation(2), we have
2x0.0413=0.05+dg/g
dg/g=0.0326
(dg/g)x100=3.26%
the percent uncertainty in the calculated value of g is 3.26%
Thank you, Could you also help me with this one?
A small block travels up a frictionless incline that is at an angle of 30.0° above the horizontal. The block has speed 4.21 m/s at the bottom of the incline. Assume
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acceleration on the frictionless incline a==gsin(30)=g/2
initial velocty=u=4.21 m/s
Final velocity=v=0
Using kinematics equation, v^2=u^22aS
S is the distance travelled by the block.
as v=0, u^2=2aS
S=u^2/2a=4.21^2/2(g/2)=4.21^2/g=17.7241/9.8=1.8 metre
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Pallvee
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