SOLUTION: Can you help with this? 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, fin

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Question 424391: Can you help with this?
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 ankor@dixie-net.com(22740) 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.
:
21t - 16t^2 = 6
A quadratic equation
-16t^2 + 21t - 6 = 0
Use the quadratic formula to find t
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
In this equation; x=t; a=-16; b=21 c=-6
t+=+%28-21+%2B-+sqrt%2821%5E2-4%2A-16%2A-6+%29%29%2F%282%2A-16%29+
:
t+=+%28-21+%2B-+sqrt%28441-384+%29%29%2F%28-32%29+
:
t+=+%28-21+%2B-+sqrt%2857+%29%29%2F%28-32%29+
Two solutions
t+=+%28-21+%2B+7.5498%29%2F%28-32%29+
t = %28-13.450%29%2F%28-32%29
t = .42 sec, 6 ft on the way up
and
t+=+%28-21+-+7.5498%29%2F%28-32%29+
t = %28-28.5498%29%2F%28-32%29
t = .89 sec, 6ft on the way down
:
:
Looks like this:
+graph%28+300%2C+200%2C+-1%2C+3%2C+-4%2C+10%2C+-16x%5E2%2B21x%2C+6%29+