ch04-p037

# ch04-p037 - 37. We designate the given velocity v = (7.6...

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(a) Different approaches are available, but since it will be useful (for the rest of the problem) to first find the initial y velocity, that is how we will proceed. Using Eq. 2-16, we have 22 2 10 0 2 (6.1 m/s) 2(9.8 m/s )(9.1 m) yy y vv g y v =− Δ ¡ which yields v 0 y = 14.7 m/s. Knowing that v 2 y must equal 0, we use Eq. 2-16 again but now with Δ y = h for the maximum height: 2 2 20 2 0 (14.7 m/s) 2(9.8 m/s ) g h h ¡ which yields h = 11 m. (b) Recalling the derivation of Eq. 4-26, but using v 0 y for v 0 sin θ 0 and v 0 x for v 0 cos 0 , we have 2 00 1 0, 2 yx vt g t R vt = which leads to 2/ . xy R vv g = Noting that v 0 x = v 1 x = 7.6 m/s, we plug in values and obtain R = 2(7.6 m/s)(14.7 m/s)/(9.8 m/s 2 ) = 23 m. (c) Since v 3 x = v 1 x = 7.6 m/s and v 3 y = – v 0 y = –14.7 m/s, we have 2 2 33 3 (7.6 m/s) ( 14.7 m/s) 17 m/s. v =+ = + = (d) The angle (measured from horizontal) for
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## This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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