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

1 Answer
Feb 7, 2016

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:

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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#