Define projectile particle and derive the equation $y\, = \,(\tan \,{\theta _0})x\, - \,\frac{g}{{(2\,\cos \,{\theta _0})}}{x^2}$
Define projectile particle and derive the equation $y\, = \,(\tan \,{\theta _0})x\, - \,\frac{g}{{(2\,\cos \,{\theta _0})}}{x^2}$
When an object is thrown in gravitational field of the Earth, it moves with constant horizontal velocity and constant vertical acceleration. Such a two dimensional motion is called a projectile motion and such an object is called a projectile.
Let the distance travelled by the object at time ' $\mathrm{t}$ ' using with $v_{\mathrm{o}}$ given by $x=\left(v_{0} \cos \theta_{0}\right) t$
Let the distance travelled by the projectile along the y direction be given by
$y=\left(v_{0} \sin \theta_{0}\right) t-1 / 2 g t^{2}$
From $(1)$ $\mathrm{t}=\frac{x}{v_{o} \cos \theta_{o}}$
Putting $\mathrm{t}$ value in $\mathrm{y}$ we get
$y=\left(v_{0} \sin \theta_{0}\right)\left(\frac{x}{v_{o} \cos \theta_{o}}\right)-\frac{1}{2} \mathrm{~g}\left(\frac{x}{v_{o} \cos \theta_{o}}\right)$
$y=x \tan \theta_{o}-\frac{1}{2} \frac{g}{\left(v_{o} \cos \theta_{o}\right)^{2}} \cdot x^{2}$
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- [JEE MAIN 2022]
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