A basketball player maintains a steady pace of 2.5 m/s while throwing a basketball vertically at 6.0 m/s. How far does the player advance before getting the ball back? Air resistance is negligible. I was unsure which formula to apply to this scenario. Is there any relevance to an angle? First, we determine the duration to reach peak height. The total time for the flight will be double the ascent duration. According to Newton's equations of motion: v = u + at. At the highest point, v = 0, where u is 6 m/s. Thus, the equation becomes 0 = 6 - 9.81t, leading us to t = 0.61 seconds. Therefore, the total flight time equals 1.22 seconds as the player runs towards the ball at a horizontal speed of 2.5 m/s. The distance traveled can be calculated using distance = speed × time, resulting in distance = 2.5 m/s * 1.22, yielding a final distance of 6.11m.
Answer: Her velocity magnitude (v) relative to the shore is 5.70 km/h.
Explanation:
Let Q be the speed of the boat, and P be the speed of the river flow.
R represents the resultant velocity combining boat velocity and river current.
According to vector addition using the law of triangles:
From the diagram:
P = 3.5 km/h, Q = 4.5 km/h
Therefore, her velocity magnitude relative to the shore is 5.70 km/h.
