Correct Answer - Option 3 : a parabola

**Concept:**

Equation of projectile will be of the form

y = Ax + Bx^{2}

**This is the equation of the parabola**

**Derivation:**

Let the projectile velocity be U with an angle \(\begin{array}{l} \theta \\ \end{array}\) from the horizontal

Horizontal component of velocity

\(\begin{array}{l} \mathop u\nolimits_x = u\cos \theta \\ \end{array}\)

vertical component of velocity

\(\begin{array}{l} \mathop u\nolimits_y = u\sin \theta - gt\\ \end{array}\)

\(x = u\cos \theta t\) (1)

\(y = u\sin \theta t - \frac{1}{2}g\mathop t\nolimits^2 \) (2)

substituting t from (1) in (2), we get

\(\begin{array}{l} y = x\tan \theta - \frac{{g\mathop x\nolimits^2 }}{{2\mathop u\nolimits^2 \mathop {\cos }\nolimits^2 \theta }}\\ \end{array}\)

**This is the equation of the parabola**