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January 31, 2015

January 31, 2015

Posted by **H** on Sunday, December 5, 2010 at 5:36pm.

- Pre -calculus -
**Reiny**, Sunday, December 5, 2010 at 6:02pmheight at beginning of observation:

tan 21° = h/2 ----> height = 2tan21 = .76773

height after 5 seconds:

tan 35° = h/2 ----> height = 1.40042

change in height = .63269

average vertical speed = .63269/5 = .12654 km/s

so in 15 seconds it will have risen 1.89061 km on top of the .76773 km , its original height

so new height = 2.66579

tanØ = 2.66278/2 = 1.3328955..

Ø = 53.12°

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