Question 471488
A ball is thrown upward with an initial velocity of 14 meters per second (mps)
 from a cliff that is 70 meters high. 
The height of the ball is given by the quadratic equation h = -4.9t^2 + 14t + 90 
where h is in meters and t is the time in seconds since the ball was thrown.
 Find the time that the ball will be 20 meters from the ground...rounded to the nearest tenth of a second
:
Just take the given equation, replace h with 20 and you have:
-4.9t^2 + 14t + 90 = 20
arrange it into a quadratic equation
-4.9t^2 + 14t + 90 - 20 = 0
-4.9t^2 + 14t + 70 = 0
Use the quadratic formula to find t
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 
In this equation x=t; a=-4.9; b=14; c=70
{{{t = (-14 +- sqrt(14^2-4*-4.9*70 ))/(2*-4.9) }}} 
:
{{{t = (-14 +- sqrt(196-(-1372) ))/(-9.8) }}}
:
{{{t = (-14 +- sqrt(196+1372 ))/(-9.8) }}} 
:
{{{t = (-14 +- sqrt(1568 ))/(-9.8) }}} 
Two solutions
{{{t = (-14 + 39.6)/(-9.8) }}} 
t = {{{25.6/(-9.8)}}}
t = -2.6, obviously this is not the solution
and 
{{{t = (-14 - 39.6)/(-9.8) }}} 
t = {{{(-53.6)/(-9.8)}}}
t = 5.47 ~ 5.5 seconds it will be at 20 meter
:
:
You can prove this to yourself, substitute 5.47 for t in the original equation and see that h ~ 20m