ch13-p098 - 2 98. (a) From Ch. 2, we have v 2 = v0 + 2ax ,...

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98. (a) From Ch. 2, we have 22 0 2 vv a x =+Δ , where a may be interpreted as an average acceleration in cases where the acceleration is not uniform. With v 0 = 0, v = 11000 m/s and Δ x = 220 m, we find a = 2.75 × 10 5 m/s 2 . Therefore, 52 4 2 2.75 10 m/s 2.8 10 9.8 m/s ag g §· × == × ¨¸ ©¹ . (b) The acceleration is certainly deadly enough to kill the passengers. (c) Again using 0 2 a x , we find 2 2 (7000 m/s) 7000 m/s 714 . 2(3500 m) = (d) Energy conservation gives the craft’s speed v (in the absence of friction and other dissipative effects) at altitude h = 700 km after being launched from R = 6.37 × 10 6
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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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