91Ó°ÊÓ

From the given parameters, when the gun is at rest, the gun fires the dart and it travels \(5.00 ~m\). During this time, the dart is under the effect of gravity and its initial vertical velocity is zero because the gun is held horizontally. In order to calculate the initial velocity of the dart (\(u\)), we first need to determine the time of flight (\(t\)). We can do this using the equation of vertical motion \(h = 0.5 g t^{2}\) where \(h = 1.00~m\) is the height and \(g = 9.81 ~m/s^{2}\) is the acceleration due to gravity. Solving for \(t\), we get \(t = \sqrt{\frac{2h}{g}}\). Once we find the time of flight, we can substitute it into the equation of horizontal motion \(d = u t\) where \(d = 5.00 ~m\) is the dart's horizontal distance. Solving for \(u\), we get \(u = \frac{d}{t}\). This gives us the dart's initial velocity.

Next, we need to find out how far the dart will travel horizontally, while the college student fires it from the moving gun. Here, time of flight remains the same because the vertical motion of the dart doesn't depend on the horizontal motion and the height from which the dart is fired is still the same. However, now the dart not only has the initial horizontal velocity \(u\) but also additional horizontal velocity due to the motion of the student which is \(2.00 ~m/s\) along the incline. The horizontal component of this velocity \(v_x\) is \(2.00 ~m/s * \cos(45^{\circ})\). So, the total horizontal velocity \(V_x\) is \(u + v_x\). We can then calculate the new horizontal distance \(D'\) using the equation \(D' = V_x * t\).