A hot-air balloon is released at 1:00 P.M. and rises vertically at a rate of 14 m/sec. An observation point is situated 100 meters from a point on the ground directly below the balloon (see the figure). If t denotes the time (in seconds) after 1:00 P.M., express the distance d between the balloon and the observation point in terms of t.

Question

brinazarski · Answer

I got d(t)=14t + (sqrt(10196)-14). Why is it wrong and how do I get the right answer?

anonymous · Answer

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In the drawing above, the balloon rises at 14m/s, and so after t seconds, it is 14t meters above the ground. The distance between the balloon and the observation point is the hypotenuse of the right triangle. Therefore, using Pythagoras, the distance x is given by:

x^2 = (14t)^2 + 100^2, and so 
x = SQRT[(14t)^2 + 10,000]

I'm not sure exactly where you went wrong but I suspect you haven't drawn the correct triangle diagram?

anonymous · Answer

FYI, the observation point is at the bottom left corner of the triangle, and the hypotenuse should be labelled x - sorry for the incomplete diagram!

brinazarski · Answer

Thank you!