Friday

November 21, 2014

November 21, 2014

Posted by **Michala** on Sunday, July 20, 2014 at 1:40pm.

h = -16t2 + 64t +400 where h is in feet, and t is in seconds.

To the nearest tenth of a second, how long does it take the rock to reach the ground?

To the nearest tenth of a second, how long will it take the rock to reach its maximum height?

To the nearest tenth of a foot, how high will the rock go before it starts falling down?

- College math -
**bobpursley**, Sunday, July 20, 2014 at 1:55pmHf=0=400+64t-16t^2

use the quadratic equation

a=-16, b=64, c=400

Now how high? Lets do some Gedunken.

if we find the time it takes for the rock to leave the tower, and return to the same altitute it started from

400=400+64t-16t^2 or

0=t(64-16t) the solutions are t=0, and t=4 seconds. Now the rock is at maximkum height in half that, or at t=2seconds

h=400+64t-16t^2 solve for t=2

- College math -
**bobpursley**, Sunday, July 20, 2014 at 1:58pmI would like to point out a basic misunderstanding your teacher has with falling down. The rock is falling the instant is thrown upwards, that is why is slows down until the max height.

Satellites in earth orbit, albeit at constant altitude, are constantly falling, otherwise, they would go off in space.

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