An arrow is shot into the air at an angle of \(60.0^{\circ}\) above the horizontal with a speed of \(20.0 \mathrm{m} / \mathrm{s} .\) (a) What are the \(x\) - and \(y\) -components of the velocity of the arrow \(3.0 \mathrm{s}\) after it leaves the bowstring? (b) What are the \(x\) - and \(y\) -components of the displacement of the arrow during the 3.0 -s interval?

Using the given angle (\(60^{\circ}\)) and the initial speed (\(20.0 \mathrm{m} / \mathrm{s}\)), we can find the initial components of the velocity using trigonometry (sine and cosine functions) as follows: $$ v_{0x} = v_0 \cos(60^{\circ}), \quad v_{0y} = v_0 \sin(60^{\circ})$$ where \(v_{0}\) is the initial velocity and \(v_{0x}\) and \(v_{0y}\) are its x and y components, respectively. Using the provided values and our calculators, we get $$ v_{0x} = 20.0 \mathrm{m} / \mathrm{s} \cos(60^{\circ}) = 20.0 \mathrm{m} / \mathrm{s} \times 0.5 = 10.0 \mathrm{m} / \mathrm{s} $$ $$ v_{0y} = 20.0 \mathrm{m} / \mathrm{s} \sin(60^{\circ}) = 20.0 \mathrm{m} / \mathrm{s} \times \sqrt{3} / 2 = 10.0\sqrt{3} \mathrm{m} / \mathrm{s} $$

The x-component of the velocity will remain constant throughout since there is no acceleration in the x-direction. So, \(v_{x}(3) = v_{0x} = 10.0 \mathrm{m} / \mathrm{s}\). The y-component of the velocity will be affected by the gravitational acceleration (\(g\)) and the initial y-component of the velocity (\(v_{0y}\)). We need to find the vertical velocity (\(v_{y}(3)\)) after 3 seconds using the following equation: $$ v_y = v_{0y} - gt $$ where \(t\) is the time elapsed (3 seconds in this case). Using the gravity (\(g = 9.8 \mathrm{m} / \mathrm{s}^2\)), we get $$ v_{y}(3) = 10.0\sqrt{3} \mathrm{m} / \mathrm{s} - 9.8 \mathrm{m} / \mathrm{s}^2 \times 3 \mathrm{s} = 10.0\sqrt{3} \mathrm{m} / \mathrm{s} - 29.4 \mathrm{m} / \mathrm{s} $$

To find the x and y components of the displacement after 3 seconds, we can use the following equations: $$ x = x_0 + v_{0x}t \\ y = y_0 + v_{0y}t - \frac{1}{2}gt^2 $$ where \(x\) and \(y\) are the coordinates of the arrow after 3 seconds. Since it starts from the origin, \(x_0 = y_0 = 0\). Substituting the values, we get $$ x = 10.0 \mathrm{m} / \mathrm{s} \times 3 \mathrm{s} = 30.0 \mathrm{m} \\ y = 10.0\sqrt{3} \mathrm{m} / \mathrm{s}\times 3 \mathrm{s} - 0.5 \times 9.8 \mathrm{m} / \mathrm{s}^2 \times (3)^2 \mathrm{s}^2 = 30.0\sqrt{3} \mathrm{m} - 44.1 \mathrm{m} $$ Therefore, the x- and y-components of the velocity after 3 seconds are \(10.0\mathrm{m}/\mathrm{s}\) and \((10\sqrt{3}-29.4)\mathrm{m}/\mathrm{s}\), respectively, and the x and y components of the displacement are \(30.0 \mathrm{m}\) and \((30\sqrt{3} - 44.1)\mathrm{m}\), respectively.