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?

Question 79829: 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?
Found 2 solutions by ankor@dixie-net.com, Earlsdon:
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
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?
:
Use the following equation:
-16t^ + 20t + 100 = h
:
-16t^2 + 20t + 100 = 80
:
-16t^2 + 20t + 100 - 80 = 0
:
-16t^2 + 20t + 20 = 0
:
Simplify, divide equation by -4 and you have:
+4t^2 - 5t - 5 = 0
:
Use the quadratic formula to find t: a=4; b=-5; c=-5
:
I got t = -.655; and t = +1.9 sec is the solution we want:
:
:
Check solution in original equations:
h - -16t^2 + 20x + 100
h = -16(1.9^2) + 20(1.9) + 100
h = -16(3.61) + 38 + 100
h = -57.56
h = 80.24 ~ 80 ft

Answer by Earlsdon(6294) (Show Source): You can put this solution on YOUR website!
The height (as a function of time) of an object propelled upward from an initial height of with an initial velocity of is given by:

In your problem:
m/s
m
You want to find at what value of t (time) will h (height) be 80 m. So, in the formula, you would set h(t) = 80 and solve for t.
Subtract 80 from both sides.
Solve for t using the quadratic formula: where:

Making the appropriate substitutions, you'll get:

or
or Discard the negative solution and keep the positive one.
The ball will reach a height of 80 meters about 4.9 seconds.
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