If a ball is hit with an initial velocity of 110 feet per second at an angle of 45 degrees from an initial height of 2 feet, how long will it be in the air?The following parametric equations are used to represent the location of the ball after t seconds.

Answer 1

So you do not give me the equations
and
you make me work in silly old units
oh well
u = x velocity = 110 cos 45 forever
Vi = initial up velocity = 110 sin 45

g = 32 ft/second^2 in old English units
v = Vi - g t = 110 sin 45 - 32 t
h = Hi + Vi t - (g/2)t^2 = 2 + 110 sin 45 * t - 16 t^2
when will h be zero ?
0 = 2 + 77.8 t - 16 t^2
solve quadratic
https://www.mathsisfun.com/quadratic-equation-solver.html
about 4.9 seconds
now if you want it in parametric form use x = u t
where u = 110 cos 45
so t = x /77.8
then
h = Hi + Vi t - (g/2)t^2 = 2 + 110 sin 45 * t - 16 t^2
or
h = 2 + 110 sin 45 * t - 16 t^2
or
h = 2 + 77.8 t -16 t^2
becomes
h = 2 + 77.8 (x/77.8) - 16 (x^2/77.8^2)
h = 2 + x - 16 x^2/77.8^2
that parabola gives you a quadratic for x when h = 0 then go back and get t