SOLUTION: A ball is thrown upward from the roof of a building 100 m tall with an initial velocity of 20 m/s. When will the ball reach a height of 80 m? I have come up with 5.583 seconds.

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: A ball is thrown upward from the roof of a building 100 m tall with an initial velocity of 20 m/s. When will the ball reach a height of 80 m? I have come up with 5.583 seconds.       Log On


   

Question 68718: A ball is thrown upward from the roof of a building 100 m tall with an initial velocity of 20 m/s. When will the ball reach a height of 80 m?
I have come up with 5.583 seconds. Is this the correct answer??? This is the equation that I used:
-10T^2 + 20T + 100 = 80. Is this right??? I need some advice on this one, please.
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
First, the formula for the height of an object (in meters) propelled upwards with an initial velocity of v%5B0%5D from an initial height of h%5B0%5D is given by:
h%28t%29+=+-4.9t%5E2+%2B+v%5B0%5Dt+%2B+h%5B0%5D where:
v%5B0%5D+=+20m%2Fs
h%5B0%5D+=+100m
You want to find h(t)= 80, the time at which the ball reaches 80 m. You will, of course, expect to get two such times, one of which won't make any practical sense because your initial height (100m) is above the 80m point. The second time will be meaninful because the ball will be on the way down. Let's see what comes out.
-4.9t%5E2+%2B+20t+%2B+100+=+80 Simplify and solve for t. Subtract 80 from both sides.
-4.9t%5E2+%2B+20t+%2B+20+=+0 Use the quadratic formula:t+=+%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F2a%29
In this case, a = -4.9, b = 20, and c = 20.
t+=+%28-20%2B-sqrt%2820%5E2-4%28-4.9%29%2820%29%29%29%2F2%28-4.9%29 Simplify.
t+=+%28-20%2B-sqrt%28400-%28-392%29%29%29%2F-9.8%29
t+=+%28-20%2B-sqrt%28792%29%29%2F-9.8
t+=+4.9seconds. This is the solution.
t+=+-0.83 Discard this as the time must be positive.