Horizontal Projectile

HORIZONTAL PROJECTION FROM THE TOP OF A TOWER 

Equation for Path (Trajectory):
                 
      
Suppose a body is projected horizontally with an initial velocity u from the top of a tower of height 'h' at time t=0. As there is no horizontal acceleration, the horizontal velocity remains constant throughout the motion.

Hence after time t, the velocity in horizontal direction will be v_x=u .
Let the body reach a point 'P' in time t. Let x and y be the coordinates of the body.

For the y-coordinate, after time t seconds
   \( y = \frac{1} {2}gt^2 \) \( [\therefore y = u_y t + \frac{1} {2}a_y t^2 ] \)..............(1)
For x- coordinate, after t seconds
x = ut (the horizontal velocity is constant)
   \( \Rightarrow t = x/u \) ........(2)
From Eqs. (1) and (2) we get
    \( y = \frac{1} {2}g\left( {\frac{x} {u}} \right)^2 \therefore y = \left( {\frac{g} {{2u^2 }}} \right)x^2 \)..........(3)
g and u being constants, \( \left( {\frac{g} {{2u^2 }}} \right) \) is a constant 
    If  \( \left( {\frac{g} {{2u^2 }}} \right) \)=k then y=\( kx^2 \).
This equation represents the equation of a parabola.