Final answer:
To find the distance the dolphin lands, we can analyze the horizontal and vertical components of its motion. Given the initial vertical velocity of 6.25 m/s at a launch angle of 45.0°, we can use trigonometry to find the initial vertical velocity, which is approximately 4.42 m/s. Using formulas for time and distance in projectile motion, we find that the dolphin lands approximately 1.99 meters away.
Step-by-step explanation:
To find the distance the dolphin lands, we can analyze the horizontal and vertical components of its motion.
Given the initial vertical velocity (v0y) of 6.25 m/s and the launch angle of 45.0°, we can use trigonometry to find the initial vertical velocity (v0y): v0y = v0 * sin(θ) = 6.25 m/s * sin(45.0°) = 6.25 m/s * 0.7071 ≈ 4.42 m/s.
Since the dolphin lands at the same vertical height as it started, we can use the time it takes to reach the highest point in its trajectory (tmax = v0y / g) to find the total time (t) it spends in the air. Using the formula s = v0y * t - 0.5 * g * t2 and substituting known values, we can solve for t. t = (v0y + √(v0y2 + 2gh)) / g = (4.42 m/s + √((4.42 m/s)2 + 2 * 9.8 m/s2 * 0 m)) / 9.8 m/s2 ≈ 0.45 s.
Finally, to find the horizontal distance (d) the dolphin lands, we can use the formula d = v0x * t, where v0x is the initial horizontal velocity. Since there is no horizontal acceleration, v0x remains constant throughout the motion, and it can be found using v0x = v0 * cos(θ) = 6.25 m/s * cos(45.0°) = 6.25 m/s * 0.7071 ≈ 4.42 m/s. Substituting known values, we have d = 4.42 m/s * 0.45 s ≈ 1.99 m. Therefore, the dolphin lands approximately 1.99 meters away.