HW2-phy2048-Spring-2014-solution

0 ms2t t 60 s then eq 2 11 applies to motion

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Unformatted text preview: gns of the components imply the vector is in the fourth quadrant. 4P17. We find t by applying Eq. 2- 11 to motion along the y axis (with vy = 0 characterizing y = ymax ): 0 = (12 m/s) + (−2.0 m/s2)t ⇒ t = 6.0 s. Then, Eq. 2- 11 applies to motion along the x axis to determine the answer: vx = (8.0 m/s) + (4.0 m/s2)(6.0 s) = 32 m/s. Therefore, the velocity of the cart, when it reaches y = ymax , is (32 m/s)^. i 1 PHY2048 Spring 2014 HW2 4P23. (a) From Eq. 4- 22 (with θ0 = 0), the time of flight is 2h 2(45.0 m) t= = = 3.03 s. g 9.80 m/s 2 (b) The horizontal distance traveled is given by Eq. 4- 21: Δx = v0t = (250 m...
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This document was uploaded on 03/24/2014.

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