b) A water balloon is fired at 34.0 m/s from a water cannon, which is aimed at an angle of 180 above the ground. The centre of the cannon's target (which has a radius of 1.0 m) is painted on the asphalt 42.0 m away from the water cannon. 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: V= 34.0 m/s 0= 18 0 d= 42.0 m Solution: i) Horizontal component: V =34cos18 =32.3m/s ii) Vertical component: V =34 sin18 =10.5m/s Then, V, =V+a. At 0 =10.5+ -9.8)(At) At 9.8 At =1.07 sec > Time required to reach max height: 1.07 s Time of flight: 2 x 1.07= 2.14 s Horizontal distance d=V x At d= 32.3 2.14