At any instant t, the vertical distance Y and | vedclass

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(C) Given equations are $2Y = 6t - gt^2$ and $X = 4t$.First, simplify the vertical displacement equation: $Y = 3t - \frac{1}{2}gt^2$.The horizontal ve

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$5$

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(C) Given equations are $2Y = 6t - gt^2$ and $X = 4t$.First, simplify the vertical displacement equation: $Y = 3t - \frac{1}{2}gt^2$.The horizontal velocity component is $V_x = \frac{dX}{dt} = \frac{d}{dt}(4t) = 4 \,m/s$.The vertical velocity component is $V_y = \frac{dY}{dt} = \frac{d}{dt}(3t - \frac{1}{2}gt^2) = 3 - gt$.At the initial instant $t = 0$, the vertical velocity is $V_{y0} = 3 - g(0) = 3 \,m/s$.The initial velocity $V_i$ is given by the magnitude of the velocity vector at $t = 0$: $V_i = \sqrt{V_x^2 + V_{y0}^2}$.Substituting the values: $V_i = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \,m/s$.

At any instant $t$, the vertical distance $Y$ and horizontal distance $X$ of a projectile are given by $2Y = 6t - gt^2$ and $X = 4t$. The initial velocity of the projectile is ($X$ and $Y$ are in $m$ and $t$ is in $s$). (in $\,m/s$)

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