A baseball player throws baseball at a velocity 139ft per second at an angle of 20 degrees. The ball leaves the players hand at a height of 5 feet. The path of the ball has a parametric equation x=v(o)t cos(o) and y-h+v(o)tsin(o)-16t^2. Write the parametric equations that describe the motion of the ball as a function of time. How long does it take for the ball to hit the ground? When is the ball at its maximum height? What is the maximum height of the ball?

Answer 1

typo
you mean y=h+v(o)tsin(o)-16t^2

Vi = initial speed up = 139 sin 20
= 47.54 ft/s
u = horizontal speed = 139 cos 20
= 131 ft/s

x = 131 t
y = 5 + 47.54 t - 16 t^2

when is y = 0?
16 t^2 - 47.54 t - 5 = 0

t = [ 47.54 +/- sqrt(2260+320)]/32

use positive time
t = 3.07 seconds in the air

Now I suppose you could use completing the square to find the vertex of that parabola giving the max height and time thereof. However it is easier to use either physics or calculus

when is the up velocity = 0? That is the top.
dy/dt = up velocity
= 0 = 47.54 - 32 t
= 1.49 seconds to stop at top

then y at top =
5+47.54(1.49)-16(1.49)^2
= 40.3 ft

Answer 2

by the way surely they ask the range, x
x = u * t in air
= 131 * 3.07 = 402 ft
That is a serious throw