Question

a particle is projected with velocity of 10m/s at an angle of elevation 60 determine i) equation of its path
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Answer 1

Given:

Particle is projected with velocity of 10m/s at an angle of elevation 60°.

To find:

Equation of path

Calculation:

Distance along x axis :

$x = 10 \cos( {60}^{ \circ} ) t \: \: \: ......(1)$

Distance along y axis:

$y= 10 \sin( {60}^{ \circ} ) t - \dfrac{1}{2} g {t}^{2} \: \: \: ......(2)$

Putting value of t in eq. (2):

$= > y= 10 \sin( {60}^{ \circ} ) \bigg \{ \dfrac{x}{10 \cos( {60}^{ \circ} ) } \bigg \} - \dfrac{1}{2} g { \bigg \{ \dfrac{x}{10 \cos( {60}^{ \circ} ) } \bigg \} }^{2}$

$= > y= x\tan( {60}^{ \circ} ) - \dfrac{g {x}^{2} }{2 {(10)}^{2} { \cos}^{2} ( {60}^{ \circ} )}$

$= > y= x \sqrt{3} - \dfrac{g {x}^{2} }{200 \times \frac{1}{4} }$

$= > y= x \sqrt{3} - \dfrac{g {x}^{2} }{50}$

$= > y= x \sqrt{3} - \dfrac{10 \times {x}^{2} }{50}$

$= > y= x \sqrt{3} - \dfrac{{x}^{2} }{5}$

So, final answer is:

$\boxed{ \bf{y= x \sqrt{3} - \dfrac{{x}^{2} }{5} }}$