SOLUTION: 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=

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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. THANK YOU!!!!!

Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
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+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
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.