91Ó°ÊÓ

As the projectile motion is horizontal, we know that the time taken to reach the point (\(x\), \(y\)) would be equal to \(t = \frac{x}{v_{0}}\), where \(v_{0}\) represents the initial velocity. Since the force acting on the bullet is exclusively the force of gravity, we can model the motion in the vertical direction as free fall. Given this, we know from the laws of motion that the formula linking displacement, initial velocity, time, and constant acceleration is \(d = vt + 0.5at^{2}\). Thus, \(y = v_{y0}t - \frac{1}{2}gt^{2}\) (where \(g\) is the acceleration due to gravity, \(t\) is the time, \(v_{y0}\) is the initial vertical velocity which is 0 in this case so we can disregard it). This simplifies to \(y = -\frac{1}{2}gt^{2}\). Substituting \(t = \frac{x}{v_{0}}\) into the equation gives us \(y = - \frac{1}{2} g (\frac{x}{v_{0}})^{2}\), which is \(y = A x^{2}\) where \(A = - \frac{1}{2} g (\frac{1}{v_{0}})^{2}\).

Using the general formula derived in step 1 for the position of the pellet, \(A = - \frac{1}{2} g (\frac{1}{v_{0}})^{2}\), this is the constant \(A\) expressed in terms of the initial (muzzle) velocity and the free-fall acceleration.

From the given problem, \(x = 3.00 \, m\) and \(y = 0.210 \, m\), we can rearrange our equation \(y = Ax^{2}\) as follows \(A = \frac{y}{x^{2}}\), substituting the given values into this equation, we get \(A = \frac{0.210 \, m}{(3.00 \, m)^{2}} = 0.0233 \, s^{2}/m\). Now, equating this to the equation for \(A\) derived in Step 2 and solving for \(v_{0}\), we can say \(-\frac{1}{2} g (\frac{1}{v_{0}})^{2} = 0.0233 \, s^{2}/m\), which gives \(v_{0} = \sqrt{\frac{g}{2A}}\). Substituting \(A = 0.0233 \, s^{2}/m\) and \(g = 9.8 \, m/s^{2}\) into the equation we get \(v_{0} = \sqrt{\frac{9.8}{2*0.0233}} \approx 14.3 \, m/s\).