What is Equation of path of a projectile when Angle of projection 45 degree ???

Answers (1)

As    y= x \tan \theta\: - \: \frac{gx^{2}}{2u^{2}\cos ^{2}\theta }

So putting  \Theta =45 ^0\\ tan\theta =1\\ and \ cos \theta =\frac{1}{\sqrt{2}}

So the equation will be

\\y= x \tan \theta\: - \: \frac{gx^{2}}{2u^{2}\cos ^{2}\theta }\\ y= x *1\: - \: \frac{gx^{2}}{2u^{2} *\frac{1}{2} }\\ y=x-\frac{gx^2}{u^2}

 

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