The path followed by a body projected along \(y\)-axis is given as by \(y = \sqrt 3 x - (1/2){x^2}\), if \(g = 10 m/s\), then the initial velocity of projectile will be – (\(x\) and \(y\) are in m)
Practice this Projectile Motion question from Engineering Entrance Exam.
Question Details
Physics
Projectile Motion
Engineering Entrance Exam
Question
The path followed by a body projected along \(y\)-axis is given as by \(y = \sqrt 3 x - (1/2){x^2}\), if \(g = 10 m/s\), then the initial velocity of projectile will be – (\(x\) and \(y\) are in m)
Solution
By comparing the coefficient of \(x^2\) in given equation with standard equation \(y = x\tan \theta - \frac{{g{x^2}}}{{2{u^2}{{\cos }^2}\theta }}\)
\(\frac{g}{{2{u^2}{{\cos }^2}\theta }} = \frac{1}{2}\)
Substituting \( \theta = {60^o}\) we get \(u=2\sqrt {10} m/s\)