Consider a particle thrown up at an angle q to the horizontal with a velocity u. The velocity of projection can be resolved into two component u cos θ and u sin θ along OX and OY respectively.
The horizontal component remains constant while the vertical component is affected by gravity.
Let P(x, y) be the position of the projectile after ‘t’ seconds. Then distance travelled ‘x’ along OX is given by,
x = horizontal component of the velocity × time
x = ucos θ × t .......... (1)
The distance travelled along the vertical direction in same time ‘t’ is,
y = (usinθ) t - \(\frac {1}{2}\)gt2 ........ (2)
Substituting, t = \(\frac {x}{u cos\theta}\) from (1) in (2),
y = ax - bx2 ...... (3) where
a = tan θ, b = \(\frac {y = gas - bx^2}{2u^2 cos^2\theta},\) are constants for given values of θ, u and g.
Equation (3) represents a parabola. Therefore trajectory of a projectile is a parabola.
