Wednesday

August 20, 2014

August 20, 2014

Posted by **ibo** on Wednesday, September 29, 2010 at 2:52pm.

- calculus -
**Reiny**, Wednesday, September 29, 2010 at 4:49pmmake a diagram.

you should have a right-angled triangle, where the angle of elevation is Ø, the adjacent is 100 and the opposite is h, the height of the rocket.

tanØ = h/100

h = 100tanØ

dh/dt = 100sec^2 Ø (dØ/dt)

when Ø = 60° or π/3 rad

dØ/dt = 12°/sec or 12π/180 rad/sec = π/15 rad/sec

sub into the dh/dt equation and you are done

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