Question

show that the trajectory of an object thrown at certain angle with the horizontal is a parabola

Answer 1

so heres the answer........

Answer 2

Trajectory of a projectile is a parabola:

• Consider a body projected with an initial velocity u at an angle  $\alpha$ with the horizontal from O.
• The body moves along OAB and reaches the ground.
• The path OAB is called as the trajectory of a projectile.

            x = horizontal displacement

            y = vertical displacement

• 'u' can be resolved into two components

            $u_{x}$=horizontal component of velocity which is constant

                                   = $ucos\alpha$

            $u_{y}$= vertical component of velocity

                                   = $usin\alpha$

• Similarly, the components of acceleration are

            $a_{x}$= horizontal component of acceleration = 0

            $a_{y}$= vertical component of acceleration = ± g

• Let 't' be the time period
• Horizontal displacement is given by

            $x=(ucos\alpha )t\\t=\frac{x}{ucos\alpha } ----------------(1)$

• From equations of motion, we have

            $s = ut + \frac{1}{2}at^{2}  \\s=y, u=u_{y}=usin\alpha , a=-g\\y=usin\alpha (\frac{x}{ucos\alpha } )-\frac{1}{2}g(\frac{x}{ucos\alpha }  )^{2} \\y=(tan\alpha )x-\frac{g}{2u^{2}cos^{2}\alpha   } x^{2} \\y=Ax-Bx^{2} \\where,\\A=tan\alpha, B=\frac{g}{2u^{2}cos^{2}\alpha}$

∴ Trajectory of a projectile is a parabola