Derive the equation of path of a projectile and hence show that equation of path of projectile is a parabolic curve.
Solution
Let us assume that a projectile is fired with an initial velocity u at an angle θ with the horizontal. Let t be the time of flight. Let x be the horizontal displacement and y be the vertical displacement.
HORIZONTAL MOTION :
In the horizontal direction,the projectile moves with a constant velocity.
Horizontal component of initial velocity u is u.cosθ
Displacement = velocity x time
x = u.cosθ x t
`t=x/(ucosθ)`
VERTICAL MOTION OF PROJECTILE:
In the vertical motion,the projectile moves under gravity and hence this is an accelerated motion.
Vertical component of initial velocity u = u.sinθ Using kinematics equation :
`s= u_yt+1/2 x a x t^2`
`y=usinθ xx x/(ucosθ)-1/2xx g xx (x/(uosθ))^2`
`y=xtanθ-(gx^2)/(2u^2 cos^2θ)`
This is the equation of the projectile
This equation is also the equation of a parabola
Thus, proved that path traced by a projectile is a parabolic curve.
