Question 424393
a ball is thrown, its height in feet h after t seconds is given:
h=vt-16t^2
v=initial upward velocity in feet per second. if v=21 feet per second, find all values of t which h=6 feet. Do not round intermediate steps, round answer 2 decimals.

..

equation:
6=21t-16t^2
-16t^2+21t-6=0
16t^2-21t+6=0
solve by quadratic formula,
a=16, b=-21, c=6

{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}

t=(-(-21)+-sqrt(21^2-4*16*6))/2*16
 =(21+-sqrt(441-384))/32
 =(21+-sqrt(57))/32
 =(21+sqrt(57)/32 or (21-sqrt(57)/32
 =0.89 sec or 0.42 sec
ans:
The ball reaches a height of 6 feet after 0.42 sec on its way up and returns to a height of 6 feet after 0.89 sec on its way down.