*******please Check My Asnwer! :'(******** Q40. A boy stands at the edge of a bridge 20.0 m above a river and throws a stone straight down with a speed of 12.0 m/s. He throws another pebble straight upward with the same speed so that it misses the edge of the bridge on the way back down and falls into the river. For each stone find (a) the velocity as it reaches the water and (b) the average velocity while it is in flight.Note: Ignore the affects of air resistance. (a) X= 20 m a=9.80m/s2 Vi=12.0 m/s Vf= ???? V_f^2 = V_i^2 + 2a (ΔX) V_f^2 = (12.0 m/s)^2 + 2(9.80 m/s^2) (20 m) V_f^2 = (144 m^2/s^2) + (19.6 m/s^2) (20 m) V_f^2 = (144 m^2/s^2) + (392 m^2/s^2) V_f^2 = (536 m^2/s^2) V_f = 23.15167381 m/s V_f = 23 m/s Therefore the velocity of the stone as it reaches the water is approximately 23 m/s. Since the pebble was thrown straight upward with the same speed so that it misses the edge of the bridge on the way back down and falls into the river, I know that if the pebble is projected upwards in a vertical direction, then the velocity at which it is projected is equal in magnitude and opposite in sign to the velocity it has when it returns to the same height. That is, a pebble projected vertically with an upward velocity of +12 m/s will have a downward velocity of –12 m/s when it returns to that same height. So the velocity of the pebble as it reaches the water is –12 m/s.