A crate of mass 1500kg is at the top of a frictionless ramp. The slope of the ramp is 15degrees to the horizontal and is 75 metres long. The weight is about to slide down the ramp, what is the velocity at the bottom of the ramp?
"m" is the mass of the object (which isn't really relevant here since the two "m"s in the formula will cancel), "g" is the acceleration due to gravity, at the surface of the earth, about 9.81 meters per second per second. "h" is the vertical height between the initial position of the object and the last position (the "opposite side" of a right triangle with hypotenuse the length of the ramp). v is the velocity (in meters per second if you use g= 9.81) of the object at the bottom of the ramp (what you are asked to find).
$\displaystyle mgh = \frac{mv^2}{2}$
basic algebra solving for $\displaystyle v$ ...
$\displaystyle v = \sqrt{2gh}$
using right triangle trig ...
$\displaystyle h = 75\sin(15^\circ) \, meters$
$\displaystyle g = 9.8 \, m/s^2$
... finish it.
Note that you didn't provide an equation to solve the given problem in the first place. Understand that when someone posts a problem in the advanced applied math forum, folks here are going to assume that the poster has the minimum requisite knowledge to at least attempt setting up their own equation based on the problem statement.