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January 29, 2015

January 29, 2015

Posted by **Mehak** on Wednesday, March 16, 2011 at 3:52pm.

The answer is suppose to be 464 m/s and please provide the whole procedure .Thanks.

- Calcuslus-Calculus-Calculus -
**Reiny**, Wednesday, March 16, 2011 at 4:01pmDid you make a diagram?

let the height be h m

let the distance between rocket and observer be d m

then

d^2 = h^2 + 2000^2

2 d dd/dt = 2h dh/dt = 0

dd/dt = (h/d) dh/dt

when h=5000 from first equation

d = 5385.165

dd/dt = (5000/5385.164)(5000(.1)) = 464.238

- Calcuslus-Calculus-Calculus -
**Mehak**, Wednesday, March 16, 2011 at 4:07pmThank you and yea i tried it but i was makin a minor mistake and now i see it .

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