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 vx=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.