A body is projected horizontally from the top | vedclass

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Question

(C) Given: Initial horizontal velocity $u_x = 20 \ ms^{-1}$,initial vertical velocity $u_y = 0 \ ms^{-1}$,acceleration $g = 10 \ ms^{-2}$,time $t = 1

Vedclass · Accepted Answer

$A-IV, B-II, C-I, D-III$

Vedclass · Answer

(C) Given: Initial horizontal velocity $u_x = 20 \ ms^{-1}$,initial vertical velocity $u_y = 0 \ ms^{-1}$,acceleration $g = 10 \ ms^{-2}$,time $t = 1 \ s$.$A$. Velocity of the body after $1 \ s$:$v_x = u_x = 20 \ ms^{-1}$$v_y = u_y + gt = 0 + 10(1) = 10 \ ms^{-1}$Resultant velocity $v = \sqrt{v_x^2 + v_y^2} = \sqrt{20^2 + 10^2} = \sqrt{500} \approx 22.4 \ ms^{-1}$. Thus,$A-IV$.$B$. Horizontal displacement after $1 \ s$:$x = u_x t = 20 	imes 1 = 20 \ m$. Thus,$B-II$.$C$. Vertical displacement after $1 \ s$:$y = u_y t + \frac{1}{2}gt^2 = 0 + \frac{1}{2}(10)(1)^2 = 5 \ m$. Thus,$C-I$.$D$. Vertical velocity after $1 \ s$:$v_y = u_y + gt = 0 + 10(1) = 10 \ ms^{-1}$. Thus,$D-III$.Therefore,the correct matching is $A-IV, B-II, C-I, D-III$.

List-$I$ (Kinematic Variable)	List-$II$ (Value)
$A$. Velocity of the body after $1 \ s$ (in $ms^{-1}$)	$I$. $5$
$B$. Horizontal displacement of the body after $1 \ s$ (in $m$)	$II$. $20$
$C$. Vertical displacement of the body after $1 \ s$ (in $m$)	$III$. $10$
$D$. Vertical velocity of the body after $1 \ s$ (in $ms^{-1}$)	$IV$. $22.4$

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