Question paper physics

Dear Student,

Projectile Given Angular Projection

• Equation of path of projectile − Suppose at any time t, the object is at point P (x, y).

For motion along horizontal direction, the acceleration a_x is zero. The position of the object at any time t is given by,

$x=x_0+u_{x}t+\frac{1}{2}{a}_x{t}^2 .............\left(i\right)$

Here, x₀ = 0, u_x = u cos θ, a_x = 0

[Velocity of an object in the horizontal direction is constant]

Putting these values in equation (i),

$x=ut\cos \theta +\frac{1}{2}\left(0\right)t^2$

⇒ x = ut cos θ

⇒ $t=\frac{x}{u\cos \theta } .................\left(ii\right)$

For motion along vertical direction, the acceleration a_y is −g.

The position of the object at any time t along the vertical direction is given by,

$y=y_0+u_{y}t+\frac{1}{2}{a}_y{t}^2 .................\left(iii\right)$

Here, $y_0=0 ,u_y=u\sin \theta , a_y=-g$

$y=u\sin \theta t+\frac{1}{2}\left(-g\right)t^2$

 $$\begin{gathered} \Rightarrow \\ y=u\sin \theta t-\frac{1}{2}g{t}^2 \end{gathered}$$

Putting the value of t from equation (ii),

$y=u\sin \theta \left(\frac{x}{u\cos \theta }\right)-\frac{1}{2}g{\left(\frac{x}{u\cos \theta }\right)}^2$

 $$\begin{gathered} \Rightarrow \\ y=x\tan \theta -\left(\frac{1}{2}\frac{g}{u^2{\cos }^2\theta }\right)x^2 \end{gathered}$$

This is an equation of a parabola. Hence, the path of the projectile is a parabola.

Regards