A ball starts 1.175 meters off the ground and shot at an angle of 40 degrees with a velocity of 3.2 m/s. Where does it land?

Answer 1

u = horizontal velocity = 3.2 cos 40 forever

Vi = initial up velocity component = 3.2 sin 40

v = Vi - 9.8 t

h = 1.175 + Vi t - 4.9 t^2
at ground, h = 0
so
4.9 t^2 - Vi t - 1.175 = 0
solve quadratic for t, use positive solution
then
range = u t