HW8-phy2048-Spring-2014-solution

4 22 gives the time of flight as t 2hg then eq 4 21

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Unformatted text preview: rt (a) implies v0 = Rω = 6.3 m/s, and we see from the figure that (with these positive direction choices) its components are v0 x = v0 cos 30° = 5.4 m s v0 y = v0 sin 30° = 3.1 m s. The projectile motion equations become 1 x = v0 x t and y = v0 y t + gt 2 . 2 We first find the time when y = H = 5.0 m from the second equation (using the quadratic formula, choosing the positive root): t= 2 −v0 y + v0 y + 2 gH g = 0.74s. Then we substitute this into the x equation and obtain x = ( 5.4 m s ) ( 0.74 s ) = 4.0 m. 2 PHY2048 Spring 2014 HW8 11P14. To find the center of mass speed v on the plateau, we use the projectile motion equations of Chapter 4. With voy = 0 (and using “h” for h2) Eq. 4- 22 gives the time- of- flight as t = 2h/g . Then Eq. 4- 21...
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