This preview shows page 1. Sign up to view the full content.
Unformatted text preview: 3.90: As in the previous problem, the horizontal distance x in terms of the angles is gx 1 tan = tan( + ) - 2 2v cos 2 ( + ) . 0
2 Denote the dimensionless quantity gx / 2v0 by ; in this case (9.80 m/s 2 )(60.0 m)cos 30.0 = 0.2486. 2(32.0 m/s) 2 The above relation can then be written, on multiplying both sides by the product cos cos( + ), cos sin cos( + ) = sin( + ) cos - , cos( + ) and so cos sin( + ) cos - cos( + ) sin = . cos( + ) = The term on the left is sin(( + ) - ) = sin , so the result of this combination is sin cos( + ) = cos . Although this can be done numerically (by iteration, trial-and-error, or other methods), the expansion sin a cos b = 1 (sin(a + b) + sin( a - b)) allows the angle to be isolated; 2 specifically, then 1 (sin(2 + ) + sin( - )) = cos , 2 with the net result that sin( 2 + ) = 2 cos + sin . a) For = 30, and as found above, = 19.3 and the angle above the horizontal is + = 49.3. For level ground, using = 0.2871, gives = 17.5. b) For = -30, the same as with = 30 may be used (cos 30 = cos(-30)) , giving = 13.0 and + = -17.0. ...
View Full Document