Calculating Work and Velocity on a Water Slide

[vxr]

Homework Statement

A person of mass ##m = 75## kg slides a distance ##d = 5## m on a straight water slide, dropping through a vertical height ##h = 25## m. Determine the mechanical work done by gravity on the person? What is the height h if the mechanical work done by the gravity is ##W = 2010## J?

Relevant Equations

Probably some of these:
##E = mgh + \frac{1}{2} mv^2##
##W = \Delta \frac{1}{2} mv^2##

I don't really understand if the initial horizontal velocity is 0? Or do I assume it's some constant? Putting aside vertical velocity.

Also how should the "mechanical work done by gravity" be calculated? Is it just ##W = \frac{1}{2}mv^2_{final} - \frac{1}{2}mv^2_{initial}##

 

[kuruman]

The mechanical work done by gravity is ##m\vec g \cdot \vec d## where ##\vec d ## is the displacement vector.

 

[BvU]

What does ##d=5## m mean ? in the context of dropping through 25 m !

 

[vxr]

I am really unsure. I tried to draw it and I came up with this:

 

[kuruman]

The statement of the problem makes one believe that the slide is an inclined plane of 5 m hypotenuse and vertical side 25 m. This is impossible unless the vertical side is something else, 2.5 m perhaps?

 

[sojsail]

Perhaps it should have said

A person of mass m=75 kg slides a horizontal distance d=5 m on a straight water slide, while dropping through a vertical height h=25 m.

?? That would mean the hypotenuse is sqrt(5^2 + 25^2) m.

 

[haruspex]

Perhaps it should have said ?? That would mean the hypotenuse is sqrt(5^2 + 25^2) m.

That would make it far taller and steeper than any I've seen.
More likely the two numbers are swapped over: 5m vertical, 25m hypotenuse.