A body of mass 10 kg is projected at an angl | vedclass

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(D) The equation of trajectory is $y = x 	an 	heta - \frac{gx^2}{2u^2 \cos^2 	heta}$.Given $	heta = 45^{\circ}$,$x = 20$,$y = 10$,and $g = 10 \ m/

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$100\sqrt{2} \hat{i} + (100\sqrt{2} - 200) \hat{j}$

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(D) The equation of trajectory is $y = x 	an 	heta - \frac{gx^2}{2u^2 \cos^2 	heta}$.Given $	heta = 45^{\circ}$,$x = 20$,$y = 10$,and $g = 10 \ m/s^2$.$10 = 20(1) - \frac{10(20)^2}{2u^2(1/2)} \Rightarrow 10 = 20 - \frac{4000}{u^2} \Rightarrow \frac{4000}{u^2} = 10 \Rightarrow u^2 = 400 \Rightarrow u = 20 \ m/s$.The time of flight $T = \frac{2u \sin 	heta}{g} = \frac{2(20)(1/\sqrt{2})}{10} = 2\sqrt{2} \ s$.At time $t = \frac{T}{\sqrt{2}} = \frac{2\sqrt{2}}{\sqrt{2}} = 2 \ s$.The velocity components are $v_x = u \cos 	heta = 20(1/\sqrt{2}) = 10\sqrt{2} \ m/s$ and $v_y = u \sin 	heta - gt = 20(1/\sqrt{2}) - 10(2) = 10\sqrt{2} - 20 \ m/s$.The momentum vector $\vec{p} = m\vec{v} = 10(10\sqrt{2} \hat{i} + (10\sqrt{2} - 20) \hat{j}) = 100\sqrt{2} \hat{i} + (100\sqrt{2} - 200) \hat{j} \ kg \cdot m/s$.

Column-$I$	Column-$II$
$(1)$ Angle of projection for a projectile launched horizontally with constant speed	$(a)$ $0$
$(2)$ Horizontal component of acceleration for a projectile launched horizontally with constant speed	$(b)$ $0^o$

$P$. Time of flight	$1$. $3$
$Q$. Time when particle is moving at an angle of $53^{\circ}$ with horizontal in upwards direction	$2$. $10$
$R$. Range of the particle	$3$. $50$
$S$. Maximum height reached by the particle	$4$. $60$

$A$ body of mass $10 \ kg$ is projected at an angle of $45^{\circ}$ with the horizontal. The trajectory of the body is observed to pass through a point $(20, 10)$. If $T$ is the time of flight,then its momentum vector,at time $t = \frac{T}{\sqrt{2}}$,is. [Take $g = 10 \ m/s^2$]

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