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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 center 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 calculation.

my answer would be:
horizontal component: v1x = 34 cos18=32.3m/s
vertical component: v2x 34 sin18 = 10.5m/s

v2 = v1+a(▲t)
0=10.5+(-9.8)(▲t)
▲t=10.5/9.8
▲t=1.07s

time taken to reach max height = 1.07s
time of flight = 2 x 1.07 = 2.14s

horizontal distance
▲d = v1x *▲t
▲d = (32.3)(2.14)
▲d=69.1

∴ It will miss the target which is 42 m away from the cannon.

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anonymous
1 answer

  1. Range = Vo^2*sin(2A)/g = 34^2*sin(36)/9.8 = 69.3 m.
    Yes, you are correct.

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    henry2,

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