Include air resistance proportional to the square of theball's speed in the previous problem. Let the drag coeffcient bc=0.5, the softball radius to be 5cm and the mass to be 200g. a) Find the initial speed of the softball needed now to clearthe fence. b) for this speed, find the initial elevation angle thatallows the ball to most easily clear the fence. By how much doesthe ball now vertically clear the fence? This was the previous problem: A strong softball player smacks the ball at a height of 0.7 mabove home plate. The ball leaves the player's bat at an elevationof 35 degrees and travels toward a fence 2m high and 60m awayin center field. What must the initial speed of the softball be toclear the center field fence? Ignore air resistance. Answer: 25.4 m/s Include air resistance proportional to the square of theball's speed in the previous problem. Let the drag coeffcient bc=0.5, the softball radius to be 5cm and the mass to be 200g. a) Find the initial speed of the softball needed now to clearthe fence. b) for this speed, find the initial elevation angle thatallows the ball to most easily clear the fence. By how much doesthe ball now vertically clear the fence? This was the previous problem: A strong softball player smacks the ball at a height of 0.7 mabove home plate. The ball leaves the player's bat at an elevationof 35 degrees and travels toward a fence 2m high and 60m awayin center field. What must the initial speed of the softball be toclear the center field fence? Ignore air resistance. Answer: 25.4 m/s
Explanation / Answer
a) The initial speed of the ball is v = (2mg / CA) Where m = 200g g = 9.8 m/s2 C = 0.5 = 1.2 kg/m3 r = 5 cm A = r2 b) The maximum height reached by the ball is H = v2 / 2g