The equations of motion of a projectile are g | vedclass

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(A) The given equations are $x = 36t$ and $2y = 96t - 9.8t^2$. Dividing the second equation by $2$, we get $y = 48t - 4.9t^2$. Comparing these with th

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$\sin^{-1}(\frac{4}{5})$

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(A) The given equations are $x = 36t$ and $2y = 96t - 9.8t^2$. Dividing the second equation by $2$, we get $y = 48t - 4.9t^2$. Comparing these with the standard equations of projectile motion: $x = (u \cos 	heta)t$ and $y = (u \sin 	heta)t - \frac{1}{2}gt^2$. We identify the horizontal component of velocity as $u \cos 	heta = 36$ and the vertical component as $u \sin 	heta = 48$. To find the angle of projection $	heta$, we calculate $	an 	heta = \frac{u \sin 	heta}{u \cos 	heta} = \frac{48}{36} = \frac{4}{3}$. Since $	an 	heta = \frac{4}{3}$, we can construct a right-angled triangle with opposite side $4$ and adjacent side $3$. The hypotenuse is $\sqrt{4^2 + 3^2} = 5$. Therefore, $\sin 	heta = \frac{	ext{opposite}}{	ext{hypotenuse}} = \frac{4}{5}$. Thus, $	heta = \sin^{-1}(\frac{4}{5})$.

The equations of motion of a projectile are given by $x = 36t$ metre and $2y = 96t - 9.8t^2$ metre. The angle of projection is:

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