The height y and the distance x along the hor | vedclass

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(C) The equations of motion for the projectile are given as:$x = 6t$ $(i)$$y = 8t - 5t^2$ (ii)Comparing these with the standard equations of motion fo

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(C) The equations of motion for the projectile are given as:$x = 6t$ $(i)$$y = 8t - 5t^2$ (ii)Comparing these with the standard equations of motion for a projectile:$x = (u \cos 	heta)t$$y = (u \sin 	heta)t - \frac{1}{2}gt^2$From equation $(i)$,the horizontal component of velocity is $u_x = u \cos 	heta = 6 	ext{ m/s}$.From equation (ii),the vertical component of velocity is $u_y = u \sin 	heta = 8 	ext{ m/s}$.The magnitude of the initial projection velocity $u$ is given by:$u = \sqrt{u_x^2 + u_y^2}$$u = \sqrt{6^2 + 8^2}$$u = \sqrt{36 + 64}$$u = \sqrt{100}$$u = 10 	ext{ m/s}$.

The height $y$ and the distance $x$ along the horizontal plane of a projectile on a certain planet (with no surrounding atmosphere) are given by $y = 8t - 5t^2 \text{ m}$ and $x = 6t \text{ m}$,where $t$ is in seconds. The velocity with which the projectile is projected is (in $\text{ m/s}$)

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