Question

The path followed by a body projected along y-axis is given as by $y = \sqrt{3} x - (1 / 2) x^{2},$ if g = 10 m/s², then the initial velocity of projectile is (x and y are in m)

Moderate

Solution

On comparing the coefficient of x² in the given equation with the standard equation, we get
$y = x \tan θ - \frac{{gx}^{2}}{2 u^{2} \cos^{2} θ} \tan θ = \sqrt{3}, θ = 60^{0} \Rightarrow \frac{g}{2 u^{2} \cos^{2} θ} = \frac{1}{2}$
Substituting $θ$ = 60°, we get $u = 2 \sqrt{10} m / s .$

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