Hey guys,just these two last word problems regarding projectile motion,and SUCCE

Question

Hey guys,just these two last word problems regarding projectile motion,and SUCCESS!!! PLs help me guys

4.
(a.) In Galileo's Two New Sciences, the author states that "for elevation (angles of projection) which exceeds or falls short of 45 degrees by equal amounts, the rages are equal." Prove this statement refer to the figure.

(b.) Find an initial speed of 30 m/s and the range of 20m, find two possible elevation angles of projection.

http://i1019.photobucket.com/albums/af320/vince3lu/4PPP.jpg <--- LINK TO THE PICTURE/FIGURE

5. A baseball leaves the bat with a speed about 45 m/s which is greater than its terminal speed in air when dropped from rest 42 m/s. Calculate the magnitude of the drag force [D=b2 (* 2 is a subscript) where D is the drag force; b is the weight (mg) of the baseball about 1.4 N.] when air resistance is taken into account, the range is reduced from 179m to 72m and the max height from 78m to 48m.

Explanation / Answer

a) This figure explains that there are two angles of projection for

the same range of a projectile. This is how because

we have, R = u^2sin2 / g

here, sin2 = sin(180 - 2)

         sin2 = sin2(90 - ) by cancelling the sin on both sides we get

            = 90 -

This means the range is same for the angles of projection and (90 - )

And the range is maximum at the angle of projection 45 degrees

If = 45 degrees then the range, R = u^2/g which is maximum

b) u = 30 m/s

     R = 20 m

we have, R =  u^2sin2 / g

        u^2sin2 / g = 20

            sin2 = 20g/u^2 = 20 * 9.8 / 900 = 0.218

            2 = sin^-1(0.218) = 12.58

               = 6.29 degrees

but we know that the range is same for the angles of projection and(90- )

So (90 - ) = 90 - 6.29 = 83.71 degrees

So two possible elevation angles of projection are 6.29 and 83.71 degrees

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