1. **Problem statement:** A golf ball is struck from point A with an initial velocity of 80 ft/s. We need to find the speed at which it strikes the ground at point B and the total time of flight from A to B. 2. **Known information and assumptions:** - Initial velocity, $v_0 = 80$ ft/s - The ball follows a projectile motion path, starting at ground level (point A) and landing back on the ground (point B). - Acceleration due to gravity, $g = 32.2$ ft/s$^2$ downward. 3. **Formulas used:** - Time of flight for projectile launched and landing at same height: $$t = \frac{2 v_0 \sin \theta}{g}$$ - Speed at impact: The horizontal velocity remains constant, $v_x = v_0 \cos \theta$. - Vertical velocity at impact: $v_y = -v_0 \sin \theta$ (same magnitude, opposite direction). - Speed at impact: $$v = \sqrt{v_x^2 + v_y^2}$$ 4. **Important note:** The problem does not specify the launch angle $\theta$. Without it, we cannot find exact numerical answers. Assuming the ball is struck at an angle $\theta$, we express answers in terms of $\theta$. 5. **Calculate time of flight:** $$t = \frac{2 \times 80 \times \sin \theta}{32.2} = \frac{160 \sin \theta}{32.2} \approx 4.97 \sin \theta \text{ seconds}$$ 6. **Calculate speed at impact:** - Horizontal velocity: $$v_x = 80 \cos \theta$$ - Vertical velocity at impact: $$v_y = -80 \sin \theta$$ - Speed magnitude: $$v = \sqrt{(80 \cos \theta)^2 + (-80 \sin \theta)^2} = \sqrt{6400 \cos^2 \theta + 6400 \sin^2 \theta} = \sqrt{6400 (\cos^2 \theta + \sin^2 \theta)} = \sqrt{6400} = 80 \text{ ft/s}$$ 7. **Interpretation:** The speed at which the ball strikes the ground is equal to the initial speed, 80 ft/s, regardless of the angle, because air resistance is neglected and energy is conserved. **Final answers:** - Time of flight: $$t \approx 4.97 \sin \theta \text{ seconds}$$ - Speed at impact: $$80 \text{ ft/s}$$