Free-fall Time equations?

The other answer works for an object being dropped, or launched horizontally at an angle of 0°. So it works for the second part of your question.

The first part of your question -- when the projectile is launched at an angle other than 0° -- is more complicated. I'll try to work out the equation of time for this situation now. To make it general, I'll assume the object (a ball, for instance, with no air resistance) is thrown upward and outward at an angle θ above the horizontal, that it's thrown from a platform "c" meters above the level where it lands, and that it's launched with an initial velocity "b".

To generalize this stuff, the angle might be negative if it's thrown downward, or the distance "c" might be zero if the ball is thrown outward across a ball field, for example. So with those initial conditions specified, here we go.

Start by getting the vertical component of the initial velocity, using the formula d = b sin θ. We'll use "d", but the horizontal component of the initial velocity can be ignored from here on.

Since the object starts out going upward from the platform (for θ > 0, and ignoring the horizontal components), the displacement formula is

s = 1/2 at^2 + dt + c

where the acceleration a = -9.8 m/s/s is negative because gravity points downward. In that equation, "d" is the vertical component of the initial velocity, and "c" is the height of the launch point. Notice that if t=0, then s=c.

You want an equation for time. Here we go:

s = 1/2 at^2 + dt + c ==> 1/2 at^2 + dt + c - s = 0

Using the quadratic formula:

t = (1/a) {-d ± √[d^2 - 2a(c-s)]} (Answer)

There's your answer, in general form.

In practice, you'll never use it that way. Typically, you'll want to know how long it takes for the ball to hit the ground. In that case, the displacement s=0. If the ball is thrown on a flat ball field, then s=c=0. And if the ball is simply dropped from a platform,then s=d=0, and the time to hit the ground is

t = (1/a) √(-2ac) = √(-2c/a) (note that a<0, so the square root is positive)

which is the same result as the other part of your question.

x = 1/2 * a * t^2

where

a is the acceleration of gravity 9.81 m/sec^2

x is the distance fallen

t is the time it takes to fall

solving for t

2x = a*t^2

2x/a = t^2

t = sqr( 2x/a)