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?

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Question 89802: 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?
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
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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?
:
Let t = time in seconds
:
The 3 elements of the is problem are:
:
-4.9t^2, gravitational force downward
20t, velocity upward
100, initial height of the ball.
:
-4.9t^2 + 20t + 100 = 80
:
-4.9t^2 + 20t + 100 - 20 = 0
:
-4.9t^2 + 20t + 20 = 0
:
Use the quadratic formula to find t, a=-4.9, b=20, c=20:
:
t+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
:
t+=+%28-20+%2B-+sqrt%28+20%5E2+-+4+%2A+-4.9+%2A+20+%29%29%2F%282%2A-4.9%29+
:
t+=+%28-20+%2B-+sqrt%28400+-+%28392%29+%29%29%2F%28-9.8%29+
:
t+=+%28-20+%2B-+sqrt%28400+%2B+392+%29%29%2F%28-9.8%29+
:
t+=+%28-20+%2B-+sqrt%28792%29%29%2F%28-9.8%29+
:
t+=+%28-20+-+28.1425%29%2F%28-9.8%29+
:
Positive solution:
t+=+%28-48.1425%29%2F%28-9.8%29+
t = +4.9 seconds to reach 80 ft above the ground
:
:
check solution by substituting 4.9 for t in the original equation:
Should be close to 80 m
-4.9%284.9%5E2%29+%2B+20%284.9%29+%2B+100+ =