Question 388858
When a ball is thrown, its height in feet h after t seconds is given by the
 equation h = vt - 16t^2, where is the initial upwards velocity in feet per second.
 If v = 36 feet per second, find all values of t for which h = 19 feet.
 Do not round any intermediate steps. Round your answer to decimal places ??
:
Using the given equation, replace v with 36, and h with 19
36t - 16t^2 = 19
Arrange this as a quadratic equation
0 = 16t^2 - 36t + 19
Solve this using the quadratic equation (ax^2 + bx + c)
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 
Let x=t, a=16, b=-36, c= 19
{{{t = (-(-36) +- sqrt(-36^2-4*16*19 ))/(2*16) }}}
:
{{{t = (36 +- sqrt(1296-1216 ))/32 }}}
:
Two solutions
{{{t = (36 + sqrt(80))/32 }}}
t = 1.404508 sec
and
{{{t = (36 - sqrt(80))/32 }}}
t = .845915 sec 
:
We can say the ball is at 19 ft ~ .8 sec on the way up and ~ 1.4 sec on the way down