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.