The position of a projectile launched from th | vedclass

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Question

(C) The horizontal position is given by $x = u_x t$. Given $x = 40 	ext{ m}$ at $t = 2 	ext{ s}$,we have $40 = u_x 	imes 2$,which gives $u_x = 20 \

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$	an^{-1} \frac{7}{4}$

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(C) The horizontal position is given by $x = u_x t$. Given $x = 40 	ext{ m}$ at $t = 2 	ext{ s}$,we have $40 = u_x 	imes 2$,which gives $u_x = 20 	ext{ m/s}$.The vertical position is given by $y = u_y t - \frac{1}{2} g t^2$. Given $y = 50 	ext{ m}$ at $t = 2 	ext{ s}$ and $g = 10 	ext{ m/s}^2$,we have $50 = u_y(2) - \frac{1}{2}(10)(2)^2$.$50 = 2u_y - 20 \Rightarrow 2u_y = 70 \Rightarrow u_y = 35 	ext{ m/s}$.The angle of projection $	heta$ is given by $	an 	heta = \frac{u_y}{u_x}$.$	an 	heta = \frac{35}{20} = \frac{7}{4}$.Therefore,$	heta = 	an^{-1} \frac{7}{4}$.

The position of a projectile launched from the origin at $t=0$ is given by $\vec{r}=(40 \hat{i}+50 \hat{j}) \text{ m}$ at $t=2 \text{ s}$. If the projectile was launched at an angle $\theta$ from the horizontal,then $\theta$ is (take $g = 10 \text{ m/s}^2$)

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