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Question

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 1.0 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. 

Answer 1

Since cos(18°) = adjacent/hypotenuse then cos(18°) = v_0x/34.0 or v_0x = 34cos(18°) = 32.336 m/s

Since sin(18°) = opposite/hypotenuse then sin(18°) = v_0y/34.0 or v_0y = 34sin(18°) = 10.507 m/s

Considering ONLY the y-direction, when the water balloon reaches its maximum height, it will be traveling 0 m/s in the y-direction so (assuming it took off at zero height), using the kinematic equation of motion v_f = v₀ + at and (in the y-direction) v₀ = 10.507 m/s, v_f = 0 m/s and the acceleration in the y-direction is the acceleration due to gravity = -9.81 m/s² so:

0 = 10.507 + (-9.81)t

9.81t = 10.507

t = 10.507/9.81 = 1.07 s but the time to get back to the ground would be the same amount so the total time is 2.14 s

During the time (2.14 s) the water balloon is traveling at 32.336 m/s in the x-direction. So d = (32.336 m/s)(2.14 s) = 69.3 meters. Since the target is 42 m ± 1 m away, the balloon shoots over the target by a fair amount. If the cannon were tipped a little lower it would be able to hit the target. Example, if the angle were 10.5° (instead of 18°) then the v_0y would be 34sin(10.5°) = 6.2 m/s and t (to max height) = 0.63 and the total time is 1.26 s so the horizontal distance traveled would be 34cos(10.5°)(1.26) = 42.2 m which hits the target.