SOLUTION: suppose a toy rocket was launched from the ground straight up with initial velocity of 64ft/sec. How many seconds after launch was the rocket 48 feet above the ground

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Question 177484: suppose a toy rocket was launched from the ground straight up with initial velocity of 64ft/sec. How many seconds after launch was the rocket 48 feet above the ground
Answer by solver91311(24713) About Me  (Show Source):
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The height function for this is h%28t%29+=+-16t%5E2+%2B+v%5Bo%5Dt+%2B+h%5Bo%5D where v%5Bo%5D is the initial velocity and h%5Bo%5D is the initial height. You are given that the initial velocity is v%5Bo%5D+=+64 ft/sec and the rocket was launched from the ground so s%5Bo%5D=+0.

So, h%28t%29+=+-16t%5E2+%2B+64t+%2B+0 and you want to solve for t when h%28t%29+=+48

-16t%5E2+%2B+64t+%2B+0+=+48-16t%5E2+%2B+64t+-+48+=+0

Giving you a quadratic in t that can be solved by ordinary means. Note that you are going to get two different answers. Does that make sense? (Hint: What goes up, must come down)