Will the balloon hit the target? Justify your response with calculations that indicate where the water balloon will land and make one suggestion about how to adjust the water cannon so that the water balloon will hit the target. Given : Let lup bepositive, v, -34.0m/s ,a, =-9.8m/s',0=18 Unknown : Ad,=? Equation: , =v,sine V1x =V, CO58 -At 2 Ad, =V, At+1/2a Ad,=v.At Solve : V = Visine 34.0 m/s X sin 18 10.5m/s V1x =V, cose 34.0m/sXcos18 . 32.3m/s Ad, =Vjy At+1/2ai 0=10.5m/sAt+1/2X(-9.8m/s)& 0=10.5m/s-4.9m/s At 10.5m/s=4.9m/s-At At=2.143 s Ad =V,At 32.3m/sX2.143s 69.214m 69.214m>42.0m Therefor theballoonwillmiss thetarget . If the cannonis movedbackward 27m the water balloonwillhit thetarget. The angle could also be adjusted closer to either 90 . or 10