Posted by **Anonymous** on Saturday, October 22, 2011 at 1:18pm.

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

- physics -
**drwls**, Saturday, October 22, 2011 at 1:55pm
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.

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