(C) The horizontal position is given by $x = u_x t$. Given $x = 40\,m$ at $t = 2\,s$,we have $40 = u_x \times 2$,so $u_x = 20\,m/s$.The vertical posit

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

Class 11 Physics 2 Motion in a Plane Horizontal Projectile Motion

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

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A
$\tan^{-1}(\frac{2}{3})$
B
$\tan^{-1}(\frac{3}{2})$
C
$\tan^{-1}(\frac{7}{4})$
D
$\tan^{-1}(\frac{4}{5})$

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2 Motion in a Plane

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Horizontal Projectile Motion

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

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Two particles are projected from the same point with the same speed $u$ such that they have the same range $R$,but different maximum heights,$h_1$ and $h_2$. Which of the following is correct?

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