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## Homework Statement

A water balloon is fired at 34.0 m/s from a water cannon, which is aimed at an angle of 18° above the ground. The centre of the cannon's target (which has a radius of 1.0m) 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. Justify your choice.

## Homework Equations

:[/B]V

_{1}v= v

_{1}sinθ

V

_{1}h= v

_{1}cosθ

Δd = v

_{1}Δt + 1/2 a Δt

^{2 Δdh = VhΔt The Attempt at a Solution a) [/B] V1v = v1sinθ V1v = 34.0 m/s sin18° = 10.5 m/s Δd = v1Δt + 1/2 a Δt2 0= (10.5 m/s) Δt + 1/2 (9.8 m/s2) Δt2 0 = Δt (10.5 m/s -4.9 m/s2 Δt) Δt = 2.1s V1h = v1cosθ V1h = 34.0 m/s cos18° = 32.3 m/s (forward) Δdh = Vh Δt Δdh= (32.3 m/s [forward]) × (2.1s) Δdh = 68 m [forward] Therefore the ball will not hit the target but pass it by 26 m. Is this correct? b) For this part, I know I have to adjust the angle of the water cannon but I am not sure how to find out what angle I should use. Any help would be greatly appreciated!}