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.