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?
:
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 = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}  
:
{{{t = (-20 +- sqrt( 20^2 - 4 * -4.9 * 20 ))/(2*-4.9) }}}
:
{{{t = (-20 +- sqrt(400 - (392) ))/(-9.8) }}}
:
{{{t = (-20 +- sqrt(400 + 392 ))/(-9.8) }}}
:
{{{t = (-20 +- sqrt(792))/(-9.8) }}}
:
{{{t = (-20 - 28.1425)/(-9.8) }}}
:
Positive solution:
{{{t = (-48.1425)/(-9.8) }}}
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(4.9^2) + 20(4.9) + 100 }}} =