## Projectile equations

As we know, projectile motion follows a parabolic path and can be resolved into separate vertical and horizontal motion components. This time we are going to talk about certain properties of parabolic flight for a projectile that lands on the same level as it was launched, e.g. a projectile that is launched from ground level and lands a certain distance away at ground level. The most important thing to note about such a projectile is that the total vertical displacement is 0, no matter how it lands.

## Time of flight

Time of flight is the time interval between projectile launch and landing. To be able to calculate the time of flight, we have to understand that the projectile rises up to maximum height then falls back to ground level again, i.e. final vertical displacement is 0.

Lets use our equation of displacement:

Now lets substitute 0 for vertical displacement (s_{y}).

and factor out t the common factor:

If a product of two terms gives us 0, it means either of the terms is equal to 0. Therefore:

As we know, the vertical displacement is 0 at two points, at launch and at landing. That is why we got two values of t. We ignore the value t=0 because that is the time of launch and we take:

Therefore:

## Time to reach maximum height

The projectile rises until it reaches a maximum height halfway through its path and then starts to fall back to earth. The time it takes to reach maximum height is half the time of flight.

## Maximum height

Maximum height can then be found by using the vertical displacement equation and the time it takes to reach maximum height.

Into this equation:

subtitute:

to get:

## Horizontal Range

Horizontal range is the horizontal distance covered by the projectile from its launch to its landing. Horizontal range formula is derived from the equation of displacement.

The acceleration is 0 in the horizontal direction so the formula becomes:

Whereby t is the time of flight.

Substituting for t in the above equation we get:

We simplify the above equation using the trigonometric identity:

To get:

## Maximum horizontal range

The horizontal range of a projectile depends on both the velocity of projection and the angle of projection. The maximum horizontal range is the furthest the projectile can go horizontally from a certain velocity of projection.

Looking at the horizontal range equation:

We can see that the horizontal range is limited by `sin2θ` **since the sine of any angle can never be greater than 1**. This means that the maximum horizontal range can only be achieved if `sin2θ = 1` and this corresponds to the value of `2θ` being equal to 90°.

Therefore for maximum horizontal range, the angle of projection = 45° and

`sin90° = 1`