Aheavy ball swings on a string in a circular arc of radius 2 m. the two highest points of the ball’s trajectory are q and q ′ ; at these points the string is ±24◦ from the vertical. point p is the lowest point of the ball’s trajectory where the string hangs vertically down. the acceleration of gravity is 9.8 m/s 2 . what is the ball’s speed at the point p? neglect air resistance and other frictional forces. answer in units of m/s what is the ball’s speed at the point q? 1. 0 m/s < kvq k < kvpk 2. kvqk < 0 m/s 3. kvqk = 0 m/s 4. kvqk > kvpk 5. kvqk = kvpk

Iwould say the answer is d

Distance covered = speed x timethe distance traveled going against the stream = distance traveled with the stream.therefore it is the speed and the time that are the variables.as he travels the same distance both ways, speed x time is the same for each way.going upstream his speed is 4 kph; going downstream his speed is 6 kph.therefore we have two simultaneous equations: time upstream x 4 = time downstream x 6time upstream + time downstream = 60 minutestherefore we have the equation: time upstream = 6/10 hour, time downstream therefore = 4/10 hour.distance traveled at 4km/hr in 6/10 hour = 24/10 = 2.4km.check: distance traveled at 6km/hr in 4/10 hour = 24/10 = 2.4km.