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

d.7.5cm is your answer

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