Answer:
60 m
Explanation:
After 3 seconds of travel at 20 m/s, the projectile is 3·20 = 60 meters horizontally from the cannon.
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The vertical height after 3 seconds is 0.9 m, so the straight-line distance from cannon to target is √(60^2 +0.9^2) ≈ 60.007 meters.
Explanation:
By using v=( f )x( lambda )
v= 45 s^-1 x 3 m
Therefore v = 135 ms^-1
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