On a planet,a particle is thrown with a veloc | vedclass

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(A) Given velocity $\vec{v} = 6 \hat{i} + (20 - 4t) \hat{j}$.Comparing with $\vec{v} = v_x \hat{i} + v_y \hat{j}$,we have $v_x = 6$ and $v_y = 20 - 4t

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$P \rightarrow 2; Q \rightarrow 1; R \rightarrow 4; S \rightarrow 3$

Vedclass · Answer

(A) Given velocity $\vec{v} = 6 \hat{i} + (20 - 4t) \hat{j}$.Comparing with $\vec{v} = v_x \hat{i} + v_y \hat{j}$,we have $v_x = 6$ and $v_y = 20 - 4t$. Acceleration $a_y = -4 \ m/s^2$.$(P)$ Time of flight $(T)$: At the highest point,$v_y = 0$. So,$20 - 4t = 0 \Rightarrow t = 5 \ s$. Total time of flight $T = 2t = 10 \ s$. Thus,$P ightarrow 2$.$(Q)$ Angle with horizontal is $53^{\circ}$: $	an 53^{\circ} = \frac{v_y}{v_x} = \frac{20 - 4t}{6}$. Since $	an 53^{\circ} = \frac{4}{3}$,we have $\frac{4}{3} = \frac{20 - 4t}{6} \Rightarrow 8 = 20 - 4t \Rightarrow 4t = 12 \Rightarrow t = 3 \ s$. Thus,$Q ightarrow 1$.$(R)$ Range $(R)$: $R = v_x 	imes T = 6 	imes 10 = 60 \ m$. Thus,$R ightarrow 4$.$(S)$ Maximum height $(H)$: $H = \frac{v_{y0}^2}{2|a_y|} = \frac{20^2}{2 	imes 4} = \frac{400}{8} = 50 \ m$. Thus,$S ightarrow 3$.Correct match is $P ightarrow 2, Q ightarrow 1, R ightarrow 4, S ightarrow 3$.

$P$. 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	$4$. $60$

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