Final answer:
To calculate the vertical distance below Niagara Falls where the water's velocity vector points downward at a 30° angle, we use principles of projectile motion and trigonometry. By setting the vertical component of the velocity equal to the horizontal speed multiplied by the tangent of 30° and solving for time and vertical displacement, we can find the distance.
Step-by-step explanation:
To find the vertical distance below the edge of Niagara Falls where the velocity vector of the water points downward at a 30° angle below the horizontal, we need to use principles from projectile motion. With a horizontal speed of 3.95 m/s, the horizontal component of the velocity (Vx) remains constant due to the absence of horizontal forces. For an angle of 30° below the horizontal, we want to achieve a situation where our vertical component of the velocity (Vy) equals Vx multiplied by the tangent of 30° (tan(30°)).
Let's denote the speed of the water just before it falls as V0 and let the acceleration due to gravity be g (9.81 m/s2). The water starts with an initial vertical velocity of 0 m/s. Using the kinematic equation Vy = g × t, we can express the time t it takes to reach the point where the velocity makes a 30° angle below the horizontal as Vy / g. For our desired Vy, we have Vx × tan(30°) / g for the time. We then use this time in the vertical displacement formula Δy = V0 × t + 0.5 × g × t2 to find the vertical distance. Since our initial vertical velocity V0 is 0, the displacement simplifies to Δy = 0.5 × g × t2.
By plugging in the values, we calculate the vertical distance:
t = Vx × tan(30°) / g
Δy = 0.5 × g × t2
Calculating the actual values will yield the exact vertical distance needed for the water at Niagara Falls to point downwards at a 30° angle.