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(D) The horizontal displacement is given by $x = u_x t = 36t$,so the horizontal component of velocity is $u_x = 36 \ m/s$.The vertical displacement is

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$60$

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(D) The horizontal displacement is given by $x = u_x t = 36t$,so the horizontal component of velocity is $u_x = 36 \ m/s$.The vertical displacement is given by $y = u_y t - \frac{1}{2} g t^2 = 48t - 4.9t^2$,so the vertical component of velocity is $u_y = 48 \ m/s$.The initial velocity $u$ is given by the magnitude of the velocity vector: $u = \sqrt{u_x^2 + u_y^2}$.Substituting the values: $u = \sqrt{36^2 + 48^2} = \sqrt{1296 + 2304} = \sqrt{3600} = 60 \ m/s$.

The horizontal and vertical displacements of a projectile at time $t$ are $x=36 t$ and $y=48 t-4.9 t^2$,respectively. Initial velocity of the projectile in $m/s$ is

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