If a projectile is fired with an initial velocity of v0 meters per second at an angle α above the horizontal and air resistance is assumed to be negligible, then its position after t seconds is given by the parametric equations;
x = (v0 cos(α))t,
y = (v0 sin(α))t −(1/2) gt^2 where g is the acceleration due to gravity (9.8 m/s2).
find the equation of the parabolic path by eliminating the parameter. (round your answers to the nearest whole number.)

y = [tg α] *x - [1/2 g/ (v0 ²cos²(α))]* x²  , which describes a parabolic path

Explanation:

since  

x = (v0 cos(α))t

where x represents horizontal distance covered by the projectile

and

y = (v0 sin(α))t −(1/2) gt^2

where y represents vertical distance, we can replace the parameter t in order to have a function of coordinates, then

x = (v0 cos(α))t → x /(v0 cos(α))= t

replacing t in the equation of y

y = (v0 sin(α))t −(1/2) gt^2 = (v0 sin(α))x/(v0 cos(α)) −(1/2) g[x/(v0 cos(α))]² =

[sin(α) /cos(α)]* x - 1/2 g x²/(v0 cos(α))² = [tg α] *x - [1/2 g/ (v0 ²cos²(α))]* x²

therefore

y = [tg α] *x - [1/2 g/ (v0 ²cos²(α))]* x²

since tg α= constant=C1 ,  [-1/2 g/ (v0 ²cos²(α))]= constant=C2 , then

y = C1*x  + C2* x²

which describes a parabolic path

electrical to thermal

answer: the metamorphic rocks are formed when they are exposed to high pressure and high temperatures. this happens when one of the earth's plates comes in contact with the other plate by traveling in opposite directions, straight into each other or slipping under one another. the friction that is caused by this activity results in   melting of rocks and when it cools they are known as metamorphic rocks. this process needs a controlled amount of pressure and temperature otherwise it may turn into magma.