Calculating Error in Horizontal Velocity for Bouncing Ball Experiment

[physicssss]

Homework Statement

I slide a ball off of a ramp (the ramp is on a table) and the ball hits the ground and bounces horizontally and vertical.
I know that horizontal velocity = horizontal distance*sqrt(gravity/2*height) or d*sqrt(g/2h)
I want to know the equation for calculation error.

The expression for the error in the horizontal component of ball's velocity is:
Select one:
a. error v = v[(error d)/d + (error g)/g +(error h)/h]
b. error v/2 = v[(error d)/d + (error g)/g +(error h)/h]
c. error v/4 = v[(error d)/d + (error g)/g +(error 2h)/h]
d. error v = v[(error d)/d + (error g)/g +(error 2h)/h]
e. error v/2 = v[(error 2d)/d + (error g)/g +(error h)/h]

Homework Equations

The Attempt at a Solution

I think its a. The problem I have is the 2h. I don't know how to deal with it. If v=sqrt(g/h) then I know for sure the answer is a. but since that 2 is there I don't know if the answer is still a.

 

[tms]

If [itex]x[/itex] is a function of measured variables [itex]u, v, \ldots[/itex],
[tex]x = f(u, v, \ldots),[/tex]
then,
[tex]\sigma_x^2 \approx \sigma_u^2 \left ( \frac{\partial x}{\partial u}
\right )^2 + \sigma_v^2 \left ( \frac{\partial x}{\partial v} \right )^2
+ \ldots \;.[/tex]
You should be able to figure it out from there. Except that I think you may have written down the possible solutions incorrectly; aren't there some missing square roots?

 

[haruspex]

None of the choices seem right to me.
Don't worry about the 2 in the 2h. That's just a factor of root 2 on the whole expression. It has no relationship to the h specifically. What matters is the powers of the variables. If e.g. z = A x^myⁿ then Δz/z = m Δx/x + n Δy/y. It's just the normal product rule of differentiation.