Question

What is the general equation for determining where a projectile launched from the ground will land?

Answer 1

This is a repeat question. However, since a general expression is to be worked out so see details below.
Range `r=(2u^2 sin theta cos theta)/9.81`

Explanation:

Let the projectile be lunched with initial velocity `u` at an angle `theta` with x -axis.
Distance traveled horizontally in time of flight t gives us the Horizontal Range `r` which is due to `cos theta` component of the initial velocity of the projectile.
Assuming zero air friction, and noticing that there is no acceleration in the horizontal direction we obtain

`r=(u cos theta)t` .........(1)

To obtain time of flight `t`.
It is observed that the projectile will rise initially, attain maximum height and then fall down due to gravity.
`usin theta` is the initial velocity in the y direction, the final velocity in y direction will be `-usintheta`
Using the formula

`v=u+g t`

Take `g= 9.81 m/s^2`. Since gravity is acting against the velocity in the y direction so it is deceleration and `-` sign is used in front of g

`-usintheta=usintheta-9.81timest`

We obtain
`t=(2usintheta)/9.81`

Substituting value of t in expression (1)
`r=(u cos theta)times (2usintheta)/9.81`
`=>r=(2u^2 sin thetacos theta)/9.81`