SOLUTION: Please help with solving this problem--I'm lost. When a ball is thrown, its height in feet h after t seconds is given by the equation h=vt-16t, where v is the initial upwards ve

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: Please help with solving this problem--I'm lost. When a ball is thrown, its height in feet h after t seconds is given by the equation h=vt-16t, where v is the initial upwards ve      Log On


   

Question 79393: Please help with solving this problem--I'm lost.
When a ball is thrown, its height in feet h after t seconds is given by the equation h=vt-16t, where v 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 answers to 2 decimal places.
Thanks so much!!
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
When a ball is thrown, its height in feet h after t seconds is given by the equation h=vt-16t, where v 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 answers to 2 decimal places.
:
I think the equations would be: -16t^2 + 36t = 19
:
-16t^2 + 36t - 19 = 0
:
Find the value of t using the quadratic formula: a=-16; b=36; c=-19
t+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
:
t+=+%28-36+%2B-+sqrt%28+36%5E2+-+4%2A-16%2A-19+%29%29%2F%282%2A-16%29+
:
t+=+%28-36+%2B-+sqrt%281296+-+1216+%29%29%2F%28-32%29+
:
t+=+%28-36+%2B-+sqrt%2880+%29%29%2F%28-32%29+
:
Two solutions:
t+=+%28-36+%2B+8.94%29%2F-32
t+=+%28-27.06%29%2F%28-32%29
t = +.85 sec (At 19' on the way up)
and
t+=+%28-36+-+8.94%29%2F-32
t+=+%28-44.94%29%2F%28-32%29
t = 1.40 sec (At 19' on the way down)
:
Would look like this:
+graph%28+300%2C+200%2C+-2%2C+4%2C+-5%2C+30%2C+-16x%5E2+%2B+36x+%29+
:
Note that it shows about 19 ft at about .85 sec and 1.4 sec
:
Has this helped you find your way?