SOLUTION: The formula h=50t-5t^2 can be used to find the height in meters of a object shot upward at an initial velocity of 50 meters per second after t seconds. When will the height be 100

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Question 4263: The formula h=50t-5t^2 can be used to find the height in meters of a object shot upward at an initial velocity of 50 meters per second after t seconds. When will the height be 100 meters (on its way down) from when it was shot?
Answer by rapaljer(4671) About Me  (Show Source):
You can put this solution on YOUR website!
Given the formula h+=+50+t+-+5t%5E2, the question is when will h=100 meters.

100+=+50t+-+5t%5E2

Since this is a quadratic equation, you must set it equal to zero by moving everything either to the left side or the right side. My preference, in order to eliminate the negative coefficient of t%5E2 is to move everything to the left side. In other words, add %2B5t%5E2 and -50t to each side.

Writing this with the highest powers of x first, it gives you
5t%5E2+-+50t+%2B+100+=+0

Divide both sides of the equation by 5:
t%5E2+-+10t+%2B+20+=+0

Solve by completing the square or quadratic formula. Completing the square is easier in this case.
t%5E2+-+10t+%2B+_____+=+-20+%2B+______

Take half of the -10 (which is -5) and square (which is 25), so you must add 25 to each side of the equation.
t%5E2+-+10t+%2B+25++=+-20+%2B+25
++%28t-5%29%5E2+=+5+

Take the square root of each side of the equation:
+%28t+-+5%29+=+0+%2B-+sqrt+%285%29+

Add +5 to each side of the equation:
+t+-+5+%2B+5++=+0+%2B+5+%2B-+sqrt+%285%29+
+t+=+5+%2B-+sqrt+%285%29+

Of course the smaller of these values is the time to reach 100 meters going up, and the larger of these values is the time required to reach 100 meters coming back down. t+=+5+-+sqrt+%285%29+ is the time required to get UP to 100 meters, and t+=+5+%2B+sqrt+%285%29+ (which is about 7.2 seconds) is the time required to reach 100 meters on the way down.