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  • A projectile is given an initial velociy of <img alt="\left ( \hat{i}+2\hat{j} \right )m/s" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cdpi%7B100%7D%20%5Cleft%20%28%20%5Chat%7Bi%7D&

A projectile is given an initial velociy of \left ( \hat{i}+2\hat{j} \right )m/s , where \hat{i} is along the ground and \hat{j} is along the vertical. If g = 10m/s2 , the equation of its trajectory is :

 

 

Option: 1

4y=2x-25x^{2}


Option: 2

y=x-5x^{2}


Option: 3

y=2x-5x^{2}


Option: 4

4y=2x-5x^{2}


Answers (1)

best_answer

\\ \text{Given :} \\ \vec{u}=\hat{i}+2\hat{j} \\ \text{ From equation of motion},\\ \vec{s}=\vec{u}t+\frac{1}{2}at^{2} \\ \\ \Rightarrow \vec{r_{f}}-\vec{r_{i}}=(\hat{i}+2\hat{j})t+\frac{1}{2}(-g\hat{j})t^{2}\\\\ \Rightarrow \vec{r_{f}}-0 =t\hat{i}+2t\hat{j}-\frac{g}{2}t^{2}\hat{j} \\ \\\Rightarrow x\hat{i}+y\hat{j}=t\hat{i}+\left ( 2t-\frac{g}{2}t^{2} \right )\hat{j} \\ \\ \text{After comparing both side, we get},

x=t \ \text{and} \ y=\left ( 2t-5t^{2} \right )\\ \text{Now, put the value of t in}\ y=\left ( 2t-5t^{2} \right )\\ \\ y=2x-5x^{2}

\\\text{Note :} \text{ the above question is solved by basic method of kinematical equation motion} \\ \text{but you can also solve this question by equation of trajectory of projectile motion. } \\ \text{Hint for alternate method :}\\ \\ y= xtan\theta \left ( 1-\frac{x}{R} \right ) \\ \ \ \text{OR,} \\ y=xtan\theta -\frac{1}{2}\frac{x^{2}g}{u^{2}cos^{2}\theta }

Posted by

Sanket Gandhi

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