The trajectory of a projectile near the surfa | vedclass

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Question

(A) The standard equation of the trajectory of a projectile is given by $y = x 	an 	heta_0 - \frac{g x^2}{2 v_0^2 \cos^2 	heta_0}$.Comparing this w

Vedclass · Accepted Answer

$	heta_0 = \cos^{-1} \left( \frac{1}{\sqrt{5}} ight)$ and $v_0 = \frac{5}{3} \, ms^{-1}$

Vedclass · Answer

(A) The standard equation of the trajectory of a projectile is given by $y = x 	an 	heta_0 - \frac{g x^2}{2 v_0^2 \cos^2 	heta_0}$.Comparing this with the given equation $y = 2x - 9x^2$:$1$. $	an 	heta_0 = 2$. Since $	an 	heta_0 = \frac{	ext{opposite}}{	ext{adjacent}} = \frac{2}{1}$,the hypotenuse is $\sqrt{2^2 + 1^2} = \sqrt{5}$. Thus,$\cos 	heta_0 = \frac{1}{\sqrt{5}}$ and $\sin 	heta_0 = \frac{2}{\sqrt{5}}$.$2$. $\frac{g}{2 v_0^2 \cos^2 	heta_0} = 9$.Substituting $g = 10$ and $\cos^2 	heta_0 = \left( \frac{1}{\sqrt{5}} ight)^2 = \frac{1}{5}$:$\frac{10}{2 v_0^2 (1/5)} = 9 \implies \frac{10}{2 v_0^2 / 5} = 9 \implies \frac{25}{v_0^2} = 9 \implies v_0^2 = \frac{25}{9} \implies v_0 = \frac{5}{3} \, ms^{-1}$.Therefore,$	heta_0 = \cos^{-1} \left( \frac{1}{\sqrt{5}} ight) = \sin^{-1} \left( \frac{2}{\sqrt{5}} ight)$ and $v_0 = \frac{5}{3} \, ms^{-1}$. Comparing with options,option $A$ is correct.

The trajectory of a projectile near the surface of the earth is given as $y = 2x - 9x^2$. If it were launched at an angle $\theta_0$ with speed $v_0$,then $(g = 10 \, ms^{-2})$:

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