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(D) Given: Initial velocity $u = 20 \sqrt{2} \, m/s$,angle $	heta = 45^{\circ}$,$g = 10 \, m/s^2$.The time taken to reach the maximum height is $t =

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$10 \sqrt{5}$

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(D) Given: Initial velocity $u = 20 \sqrt{2} \, m/s$,angle $	heta = 45^{\circ}$,$g = 10 \, m/s^2$.The time taken to reach the maximum height is $t = \frac{u \sin 	heta}{g} = \frac{20 \sqrt{2} \cdot \sin(45^{\circ})}{10} = \frac{20 \sqrt{2} \cdot (1/\sqrt{2})}{10} = 2 \, s$.The horizontal displacement at maximum height is $x = u \cos 	heta \cdot t = (20 \sqrt{2} \cdot \cos(45^{\circ})) \cdot 2 = 20 \cdot 2 = 40 \, m$.The vertical displacement at maximum height is $y = H = \frac{u^2 \sin^2 	heta}{2g} = \frac{(20 \sqrt{2})^2 \cdot (1/\sqrt{2})^2}{2 \cdot 10} = \frac{800 \cdot 0.5}{20} = 20 \, m$.The total displacement vector is $\vec{s} = x \hat{i} + y \hat{j} = 40 \hat{i} + 20 \hat{j}$.The magnitude of displacement is $|\vec{s}| = \sqrt{40^2 + 20^2} = \sqrt{1600 + 400} = \sqrt{2000} = 20 \sqrt{5} \, m$.Average velocity $v_{avg} = \frac{	ext{Displacement}}{	ext{Time}} = \frac{20 \sqrt{5}}{2} = 10 \sqrt{5} \, m/s$.

$A$ stone is projected with a velocity $20 \sqrt{2} \, m/s$ at an angle of $45^{\circ}$ to the horizontal. The average velocity of the stone during its motion from the starting point to its maximum height is $.......... \, m/s$ (take $g = 10 \, m/s^2$).

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