The three equations we have used for projectile motion are;

For horizontal displacement **x = u _{x}**t (i)

For vertical displacement

For vertical velocity

Equations (i) and (ii) are variations of suvat equation 3 (s = ut + ^{1}/_{2}at^{2})

Equation (iii) is a variation of suvat equation 2 (v = u - at)

In some examples of projectile motion the projectile is fired horizontally from a higher ground level (or the top of a building etc.)

Therefore in these examples the projectile's initial vertical velocity (u_{y}) is zero

This means:

equation (ii) simplifies to **y = ^{1}/_{2}gt^{2}**

equation (iii) simplifies to

**Note** the change in sign i.e. dropping the "-" from "- ^{1}/_{2}gt^{2} " and dropping the "-" from "- gt "
and the change of direction of the y axis in the diagram above

**Explanation**

In the previous example their was a change in direction of vertical component of the projectiles velocity (v_{y}).
Therefore we used "+" for up and "-" for down.
Acceleration due to gravity (g) acts down so it would also be negative. y in the previous example represented the height __above__ the ground.

In this example the vertical component of the velocity starts at 0 and then increases in magnitude but is always directed downwards. So as there
is only one direction involved now then we just choose down to be "+"". This means g is now positive and y will now be the distance the object __drops__
from the initial height it was launched from. This is illustrated by the change in the direction of the y axis in the diagram above.

**Example**

A projectile is launched horizontally from the top of a building with an initial velocity of 10 ms^{-1}

If the object lands 30 metres from the base of the building then what is the height of the building?

- u
_{x}= 10 ms^{-1}(initial velocity is horizontal) - x = 30m
- g = 9.81 ms
^{-1}

x = u_{x}t therefore t = x/u_{x} t = 30/10 = **3s**

y = ^{1}/_{2}gt^{2} therefore y = ^{1}/_{2} x 9.81 x 3^{2} = **44.15m**