A baseball pitcher delivers a fastball that crosses the plate at an angle of \(7.25^{\circ}\) relative to the horizontal and a speed of \(88.5 \mathrm{mph}\). The ball (of mass \(0.149 \mathrm{~kg}\) ) is hit back over the head of the pitcher at an angle of \(35.53^{\circ}\) with respect to the horizontal and a speed of \(102.7 \mathrm{mph}\). What is the magnitude of the impulse received by the ball?

First, we need to convert the given speeds from mph to meters per second (m/s). To perform the conversion, we can use the following formula: Speed (m/s) = Speed (mph) × 0.44704 Initial Speed = \(88.5 \mathrm{mph} \times 0.44704 = 39.56 \mathrm{m/s}\) Final Speed = \(102.7 \mathrm{mph} \times 0.44704 = 45.91 \mathrm{m/s}\)

Next, we need to find the x and y components of the initial and final momenta. To do this, we will use trigonometry. For the initial momentum, we have: Initial momentum in x direction = mass × initial speed × cos(initial angle) Initial momentum in y direction = mass × initial speed × sin(initial angle) For the final momentum, we have: Final momentum in x direction = mass × final speed × cos(final angle) Final momentum in y direction = mass × final speed × sin(final angle) Substitute the given values and calculate the components: Initial momentum in x direction = \(0.149 \mathrm{~kg} \times 39.56 \mathrm{m/s} \times \cos(7.25^{\circ}) = 5.84 \mathrm{kg \cdot m/s}\) Initial momentum in y direction = \(0.149 \mathrm{~kg} \times 39.56 \mathrm{m/s} \times \sin(7.25^{\circ}) = 0.50 \mathrm{kg \cdot m/s}\) Final momentum in x direction = \(0.149 \mathrm{~kg} \times 45.91 \mathrm{m/s} \times \cos(35.53^{\circ}) = 5.68 \mathrm{kg \cdot m/s}\) Final momentum in y direction = \(0.149 \mathrm{~kg} \times 45.91 \mathrm{m/s} \times \sin(35.53^{\circ}) = 8.33 \mathrm{kg \cdot m/s}\)

Find the change in the x and y components of the momentum: Change in momentum in x direction = Final momentum in x direction - Initial momentum in x direction Change in momentum in y direction = Final momentum in y direction - Initial momentum in y direction Change in momentum in x direction = \(5.68 \mathrm{kg \cdot m/s} - 5.84 \mathrm{kg \cdot m/s} = -0.16 \mathrm{kg \cdot m/s}\) Change in momentum in y direction = \(8.33 \mathrm{kg \cdot m/s} - 0.50 \mathrm{kg \cdot m/s} = 7.83 \mathrm{kg \cdot m/s}\)

Finally, find the magnitude of the impulse using the Pythagorean theorem: Impulse magnitude = \(\sqrt{(Change~in~momentum~in~x~direction)^2 + (Change~in~momentum~in~y~direction)^2}\) Impulse magnitude = \(\sqrt{(-0.16 \mathrm{kg \cdot m/s})^2 + (7.83 \mathrm{kg \cdot m/s})^2} = 7.84 \mathrm{kg \cdot m/s}\) The magnitude of the impulse received by the ball is approximately \(7.84 \mathrm{kg \cdot m/s}\).