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, $u_y = 0$.
Acceleration due to gravity, $g = a_y = 10 \,ms

Vedclass · Accepted Answer

$IV, II, I, III$

Vedclass · Answer

(C) Given:
Initial horizontal velocity, $u_x = 20 \,ms^{-1}$.
Initial vertical velocity, $u_y = 0$.
Acceleration due to gravity, $g = a_y = 10 \,ms^{-2}$.
Horizontal acceleration, $a_x = 0$.
For $t = 1 \,s$:
$A$. Resultant velocity $v = \sqrt{v_x^2 + v_y^2}$.
$v_x = u_x + a_x t = 20 + 0 = 20 \,ms^{-1}$.
$v_y = u_y + a_y t = 0 + 10 	imes 1 = 10 \,ms^{-1}$.
$v = \sqrt{20^2 + 10^2} = \sqrt{400 + 100} = \sqrt{500} \approx 22.4 \,ms^{-1}$. (Matches $IV$)
$B$. Horizontal displacement $s_x = u_x t + \frac{1}{2} a_x t^2 = 20 	imes 1 + 0 = 20 \,m$. (Matches $II$)
$C$. Vertical displacement $s_y = u_y t + \frac{1}{2} a_y t^2 = 0 + \frac{1}{2} 	imes 10 	imes 1^2 = 5 \,m$. (Matches $I$)
$D$. Vertical velocity $v_y = 10 \,ms^{-1}$. (Matches $III$)
Thus, the correct matching is $A-IV, B-II, C-I, D-III$.

$A$. 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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