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?

Algebra ->  Customizable Word Problem Solvers  -> Geometry -> 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?      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

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) About Me  (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) About Me  (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 h%5B0%5D with an initial velocity of v%5B0%5D is given by:
h%28t%29+=+-4.9t%5E2%2Bv%5B0%5Dt%2Bh%5B0%5D
In your problem:
v%5B0%5D+=+20m/s
h%5B0%5D+=+100m
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.
80+=+-4.9t%5E2%2B20t%2B100 Subtract 80 from both sides.
-4.9t%5E2%2B20t%2B20+=+0 Solve for t using the quadratic formula:t+=+%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F2a where:
a+=+-4.9
b+=+20
c+=+20 Making the appropriate substitutions, you'll get:
t+=+%28-20%2B-sqrt%28%2820%29%5E2-4%28-4.9%29%2820%29%29%29%2F2%28-4.9%29
t+=+%28-20%2B-sqrt%28400%2B392%29%29%2F-9.8
t+=+%28-20%2B-sqrt%28792%29%29%2F%28-9.8%29
t+=+%28-20%2F%28-9.8%29%29%2B28.14%2F%28-9.8%29 or t+=+%28-20%2F%28-9.8%29%29-28.14%2F%28-9.8%29
t+=+-0.831 or t+=+4.913 Discard the negative solution and keep the positive one.
The ball will reach a height of 80 meters about 4.9 seconds.