A particle is projected horizontally from a t | vedclass

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Question

(A) Given:Initial horizontal velocity $u_x = 10\,m/s$.Initial vertical velocity $u_y = 0\,m/s$.Acceleration due to gravity $g = 10\,m/s^2$.Time $t = 1

Vedclass · Accepted Answer

$(A \rightarrow q, B \rightarrow q, C \rightarrow q, D \rightarrow p)$

Vedclass · Answer

(A) Given:Initial horizontal velocity $u_x = 10\,m/s$.Initial vertical velocity $u_y = 0\,m/s$.Acceleration due to gravity $g = 10\,m/s^2$.Time $t = 1\,s$.$(A)$ Horizontal component of velocity $(v_x)$: In projectile motion,horizontal velocity remains constant. So,$v_x = u_x = 10\,m/s$. Matches with $(q)$.$(B)$ Vertical component of velocity $(v_y)$: Using $v_y = u_y + g_y t$,where $g_y = g = 10\,m/s^2$. So,$v_y = 0 + 10(1) = 10\,m/s$. Matches with $(q)$.$(C)$ Horizontal displacement $(x)$: $x = u_x t = 10(1) = 10\,m$. Matches with $(q)$.$(D)$ Vertical displacement $(y)$: $y = u_y t + \frac{1}{2} g t^2 = 0(1) + \frac{1}{2}(10)(1)^2 = 5\,m$. Matches with $(p)$.Therefore,the correct matching is $(A \rightarrow q, B \rightarrow q, C \rightarrow q, D \rightarrow p)$.

Column $I$	Column $II$
$(A)$ Horizontal component of velocity	$(p)$ $5$ $SI$ unit
$(B)$ Vertical component of velocity	$(q)$ $10$ $SI$ unit
$(C)$ Horizontal displacement	$(r)$ $15$ $SI$ unit
$(D)$ Vertical displacement	$(s)$ $20$ $SI$ unit

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