A cartasian equation of a projectile is given by $\mathrm{y}=2 \mathrm{x}-5 \mathrm{x}^{2}$ Calculate its initial velocity. (A) \(\sqrt{10 \mathrm{~ms}^{-1}}\) (B) \(\sqrt{5 \mathrm{~ms}^{-1}}\) (C) \(\sqrt{2} \mathrm{~ms}^{-1}\) (D) \(4 \mathrm{~ms}^{-1}\)

Since the Cartesian equation of a projectile is given by \(y = 2x - 5x^2\), we can first find the initial vertical velocity by taking the derivative of it with respect to time. We know that the horizontal velocity remains constant throughout the projectile motion, so: \[v_x = \frac{dx}{dt}\] The vertical velocity, on the other hand, changes with time and the vertical position: \[v_y = \frac{dy}{dt} = \frac{d(2x - 5x^2)}{dt}\] To find \(v_y\), we can first find the derivative of \(y\) with respect to \(x\): \[\frac{dy}{dx} = \frac{d(2x - 5x^2)}{dx} = 2 - 10x\] Then, we can find the initial vertical velocity when x = 0: \[v_{y0} = 2 - 10(0) = 2\] And since the horizontal velocity is constant, the initial horizontal velocity is: \[v_{x0} = \frac{dx}{dt} = \frac{2}{2} = 1\]

Now that we have the initial vertical and horizontal velocities, we can use the Pythagorean theorem to find the magnitude of the initial velocity: \[v_0 = \sqrt{v_{x0}^2 + v_{y0}^2} = \sqrt{1^2 + 2^2} = \sqrt{1+4} = \sqrt{5}\] Thus, the initial velocity of the projectile is: \(\boxed{\sqrt{5}\,\text{ms}^{-1}}\), which corresponds to option (B).