A rocket is launched from the top of a 70 ft cliff with an initial velocity of 100 ft/sec. The height (h) of the rocket after t seconds is given in the equation h=-16t^2+100t+70. How long after the rocket is launched will it be 20 ft from the ground? t=? seconds

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A rocket is launched from the top of a 70 ft cliff with an initial velocity of 100 ft/sec. The height (h) of the rocket after t seconds is given in the equation h=-16t^2+100t+70. How long after the rocket is launched will it be 20 ft from the ground? t=? seconds

Mathematics
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\[20 = -16t^2 + 100t +70\] \[0 = -16t^2 + 100t + 50\] now you just need to solve the quadratic equation for t...
and think about what the negative result means in this example and how much sense it makes.
First of all, The rocket is launched from 70 feet above the ground at 100 ft per second. so you can imagine it going up and coming down. When it comes back down, and is at the same height (70 feet above the ground), it will have a velocity equal to that of the initial velocity (100 feet per second), but in the negative direction, therfore -100 ft/s). Therefore in order to get to 20 feet above the ground, you need to make the h equal to -50. Since 20ft is 50ft less than the initial height. -50 = -16t^2+100t+70. Solve using the quadratic, only use the positive result. A negative time is not possible in this case - again, I won't bother explaining. The reason MuH4hA is incorrect is because he is assuming you start at ground level. Using his equation, you would solve for the times when the rocket is 90 feet above the ground.

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so i am getting 5/8, is that correct?
Oops forgot to delete the first half of the paragraph - I was going to make it more complicated than it needed to be. Jut look at the part starting with "Therefore in order to get ..."
Ok, now I am getting 1/8
Muh4Ha is correct cause , when t=0 , Your hight is 70feet for equation , So 0=−16t^2+100t+50
so then 5/8 is the answer then?
please do not listen to "guyguyguy" - the 70 ft in the example is the last term in your equation - it's h(0). Also, the answer should be \[{5 \over 8} (5 + \sqrt{33}) \] if I'm not mistaken...
that would be about 6.72 seconds btw.
ok, I've got that. Thank you, this math class is getting the best of me!! Last week, thank goodness!!

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