The initial velocity of a particle of mass 2, | vedclass

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(A) Given: Mass $m = 2\,kg$,initial velocity $\vec{u} = (4 \hat{i} + 4 \hat{j})\,m/s$,force $\vec{F} = -20 \hat{j}\,N$.Acceleration $\vec{a} = \frac{\

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(A) Given: Mass $m = 2\,kg$,initial velocity $\vec{u} = (4 \hat{i} + 4 \hat{j})\,m/s$,force $\vec{F} = -20 \hat{j}\,N$.Acceleration $\vec{a} = \frac{\vec{F}}{m} = \frac{-20 \hat{j}}{2} = -10 \hat{j}\,m/s^2$.For $y$-motion: $u_y = 4\,m/s$,$a_y = -10\,m/s^2$,initial $y_0 = 0$. The $y$-coordinate is zero when displacement $y = 0$.Using $y = u_y t + \frac{1}{2} a_y t^2$,we get $0 = 4t - 5t^2$.Solving for $t$,$t(4 - 5t) = 0$. Since $t 
eq 0$,$t = \frac{4}{5} = 0.8\,s$.For $x$-motion: $u_x = 4\,m/s$,$a_x = 0$.The $x$-coordinate is $x = u_x t = 4 	imes 0.8 = 3.2\,m$.

The initial velocity of a particle of mass $2\,kg$ is $(4 \hat{i} + 4 \hat{j})\,m/s$. $A$ constant force of $-20 \hat{j}\,N$ is applied on the particle. Initially,the particle was at $(0,0)$. Find the $x$-coordinate of the point where its $y$-coordinate is again zero. $..........\,m$

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