quadratic equation for trajectories?

Question

who know's?

Answer 1

I think it's -gt^2 + v*(sin theta) + h
Where g is the acceleration due to gravity, v is the initial velocity, h is the height launched at, and theta is the angle from the ground.

Answer 2

If the particle is projected at a speed of u at an angle of θ to the horizontal then it has components of ucosθ horizontally and usinθ vertically

For horizontal the component of motion s = ucosθt

For vertical motion there are two quadratics involved

s = ucosθ t - ½ g t² and (vcosθ)² = (ucosθ)² - 2gs

Answer 3

for constant acceleration in the direction of travel only:
s = s0 + v0t + (1/2)at^2