A batter hits the baseball A with an initial velocity of v0 = 100 ft/sec directly toward fi elder B at an angle of 30° to the horizontal; the initial position of the ball is 3 ft above ground level. Fielder B requires 14 sec to judge where the ball should be caught and begins moving to that position with constant speed. Because of great experience, fi elder B chooses his running speed so that he arrives at the "catch position" simultaneously with the baseball. The catch position is the fi eld location at which the ball altitude is 7 ft. Determine the velocity of the ball relative to the fi elder at the instant the catch is made.

47.4 ft/s

Explanation:

Using v² = u² - 2gs where u = initial vertical velocity of ball = 100sin30, v = final vertical velocity of ball at the time it was caught by the fielder, g = acceleration due to gravity = 32 ft/s² and s = vertical displacement = y₂ - y₁, y₁ = 3 ft and y₂ = 7 ft. s = y₂ - y₁ = 7 - 3 = 4 ft

Substituting these values into the equation, we have

v² = u² - 2gs

v² = (100sin30)² - 2 × 32ft/s² × 4 ft

v² = (100sin30 ft/s)² - 256 ft²/s²

v² = (50 ft/s)² - 256 ft²/s²

v² = 2500 ft²/s² - 256 ft²/s²

v² = 2244 ft²/s²

v = √2244 ft²/s²

v = 47.37 ft/s

v ≅ 47.4 ft/s

So, the velocity of the ball relative to the fielder at the instant the catch is made is 47.4 ft/s.

the right answer is 0.3 meters per second