Asked by Anonymous
A cannon fires a shell at an angle of 20º to the horizontal. The shell reaches a range of 436 meters, landing at the same elevation as the starting point. What was the initial velocity of the shell
Answers
Answered by
drwls
Range = (Vo^2/g)sin40 = 436 m
Solve for Vo
Yes, that is sin 40, not sin 20 in the equation. That is because of the trigonometric identity
2 sinx cosx = sin(2x)
Derive the range and you will see why both sinx and cosx appear.
Solve for Vo
Yes, that is sin 40, not sin 20 in the equation. That is because of the trigonometric identity
2 sinx cosx = sin(2x)
Derive the range and you will see why both sinx and cosx appear.
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