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

Feb 7, 2016

This is a repeat question. However, since a general expression is to be worked out so see details below.
Range $r = \frac{2 {u}^{2} \sin \theta \cos \theta}{9.81}$

#### 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

$r = \left(u \cos \theta\right) t$ .........(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.
$u \sin \theta$ is the initial velocity in the y direction, the final velocity in y direction will be $- u \sin \theta$
Using the formula

$v = u + g t$

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

$- u \sin \theta = u \sin \theta - 9.81 \times t$

We obtain
$t = \frac{2 u \sin \theta}{9.81}$

Substituting value of t in expression (1)
$r = \left(u \cos \theta\right) \times \frac{2 u \sin \theta}{9.81}$
$\implies r = \frac{2 {u}^{2} \sin \theta \cos \theta}{9.81}$