```Question 104025
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?
:
This can be made into a quadratic equation with the three elements
Let t = time in seconds
1. Gravity (negative because it's accelerated downward) -4.9t^2
2. Initial velocity (positive because it's thrown up-ward: + 20t
3. Initial height above ground: +100
:
-4.9t^2 + 20t + 100 = height above ground
:
They want to know when will it be at a height of 80 ft:
-4.9t^2 + 20t + 100 = 80
:
-4.9t^2 + 20t + 100 - 80 = 0; subtracted 80 from both sides
:
-4.9t^2 + 20t + 20 = 0
:
This is the quadratic formula: except we will be solving for t instead of x:
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
Substitute: a=-4.9; b=20; c=20
{{{t = (-20 +- sqrt( 20^2 - 4 * -4.9 * 20 ))/(2*-4.9) }}}
Do the math:
{{{t = (-20 +- sqrt(400 - (-392) ))/(-9.8) }}}
{{{t = (-20 +- sqrt(400 + 392 ))/(-9.8) }}}; minus a minus is a plus
:
{{{t = (-20 +- sqrt(792))/(-9.8) }}}
:
{{{t = (-20 +- 28.1425)/(-9.8) }}}
Solution 1
{{{t = (-20 +28.1425)/(-9.8) }}}
{{{t = 8.1425/(-9.8)}}}
t = -.83, this is NOT our solution
:
Solution 2:
{{{t = (-20 - 28.1425)/(-9.8) }}}
{{{t = -48.1425/(-9.8)}}}
:
t = + 4.9 sec, this IS our solution
```