An object of mass \(3.0 \mathrm{kg}\) is projected into the air at a \(55^{\circ}\) angle. It hits the ground 3.4 s later. What is its change in momentum while it is in the air? Ignore air resistance.

To find the initial velocities, first, we have to determine the overall initial velocity (\(v_0\)) of the object using the information given. For this, we can use the formula \(v_x = v_0 \cos(55^{\circ})\) and \(v_y = v_0 \sin(55^{\circ})\). Given that the object hits the ground after 3.4 s and the vertical acceleration (\(a_y\)) is \(-9.81 \mathrm{m/s^2}\) (due to gravity). We can use the equation \(y_f = y_i + v_{y_i}t + \frac{1}{2}a_yt^2\) to find the initial vertical velocity (\(v_{y_i}\)). The initial and final positions in the vertical direction are the same (\(y_i = y_f = 0\)). Therefore, we can rewrite the equation as \(v_{y_i} = -\frac{1}{2}a_yt^2 = -\frac{1}{2}(-9.81)(3.4)^2 = 17.9 \mathrm{m/s}\). Now, we can find the overall initial velocity (\(v_0\)) using the equation \(v_0 = \frac{v_{y_i}}{\sin(55^{\circ})} = \frac{17.9}{\sin(55^{\circ})} = 21.4 \mathrm{m/s}\). Finally, we can calculate the initial horizontal velocity, \(v_{x_i} = v_0 \cos(55^{\circ}) = 21.4 \cos(55^{\circ}) = 12.3 \mathrm{m/s}\).

Since there is no air resistance, the horizontal velocity remains constant throughout the motion. Therefore, \(v_{x_f} = v_{x_i} = 12.3 \mathrm{m/s}\). Using the same equation we used for the initial vertical velocity, we can find the final vertical velocity (\(v_{y_f}\)) as \(v_{y_f} = \frac{1}{2}a_yt^2 = \frac{1}{2}(-9.81)(3.4)^2 = -17.9 \mathrm{m/s}\).

Now, we can calculate the initial and final momenta using the mass and velocities. The initial momentum (\(p_i\)) is calculated as \(p_i = m(v_{x_i}, v_{y_i}) = 3.0 \mathrm{kg} (12.3, 17.9) \mathrm{m/s}\). The final momentum (\(p_f\)) is calculated as \(p_f = m(v_{x_f}, v_{y_f}) = 3.0 \mathrm{kg} (12.3, -17.9) \mathrm{m/s}\).

Finally, we can find the change in momentum by subtracting the initial momentum from the final momentum: \(\Delta p = p_f - p_i = 3.0 \mathrm{kg} (12.3, -17.9) \mathrm{m/s} - 3.0 \mathrm{kg} (12.3, 17.9) \mathrm{m/s} = 3.0 \mathrm{kg} (0, -35.8) \mathrm{m/s}\). Hence, the change in momentum while the object is in the air is \(3.0 \mathrm{kg} (0, -35.8) \mathrm{m/s}\).