Chapter 8, Problem 1P

To determine

To check: whether the trajectory of a projectile is parabolic.

Explanation of Solution

The parametric equation of a projectile is,

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

And,

$x=\left(v_o\cos \theta \right)t$

Where,

• x is the horizontal distance covered the projectile at any time instant t.
• y is the vertical distance covered by the projectile at any time instant t
• $v_o$ is the initial velocity of the projectile.
• $\theta$ is the projection angle of the projectile.
• $g$ is the acceleration due to the gravity.

Consider the horizontal motion of the projectile.

$\begin{array}{l}x=\left(v_o\cos \theta \right)t\\ t=\frac{x}{\left(v_o\cos \theta \right)}\end{array}$ (1)

Consider the vertical motion of the projectile.

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

Substitute the value of $t$ from equation (1). in the above equation.

$\begin{array}{l}y=\frac{v_o\sin \theta \times x}{v_o\cos \theta }-\frac{1}{2}g{\left(\frac{x}{v_o\cos \theta }\right)}^2\\ y=x\tan \theta -\frac{g}{2v_o^2{\cos }^2\theta }{x}^2\end{array}$

The above equation is of the form of a parabola.

Hence the trajectories of projectiles are parabolic.