SOLUTION: please, help mw with this A ball is thrown upward at a velocity of 12 feet per second from a height that is 40 feet above the ground. The height A (in feet) of the ball at time

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equation Customizable Word Problems -> SOLUTION: please, help mw with this A ball is thrown upward at a velocity of 12 feet per second from a height that is 40 feet above the ground. The height A (in feet) of the ball at time      Log On


   

Question 136159: please, help mw with this
A ball is thrown upward at a velocity of 12 feet per second from a height that is 40 feet above the ground. The height A (in feet) of the ball at time t (in seconds), after it is thrown, can be found by the formula h = -12t2 + 14t + 40

Find the time, in seconds, when the ball is again 40 feet above the ground.

Thanks
Found 2 solutions by vleith, JSmall:
Answer by vleith(2983) About Me  (Show Source):
You can put this solution on YOUR website!
Given: h+=+-12t2+%2B+14t+%2B+40++
You are asked to find the time(s) when the ball is 40 feet above the ground.
h+=+-12t2+%2B+14t+%2B+40+
40=+-12t2+%2B+14t+%2B+40++
+0+=+-12t%5E2+%2B+14t
0+=+-2t%286t+-+7%29+
the value is 0, if either or both terms on the right are 0.
So, 0 = -2t --> t = 0
0 = (6t-7) --> t = 7/6
The ball is at 40 feet when you throw it, and again 7/6 seconds later
Check your answer
h+=+-12%287%2F6%29%5E2+%2B+14%287%2F6%29+%2B+40+
h+=+-16.333+%2B+16.333+%2B+40+
h+=+40+ check

Answer by JSmall(7) About Me  (Show Source):
You can put this solution on YOUR website!
Since you are interested in when the height is 40, set h to 40:
h = -12t2 + 14t + 40
40 = -12t2 + 14t + 40
Now solve this equation for t by
1) Getting one side equal of the equation to zero; and then
2) Using factoring or the quadratic formula to solve for t
To get one side equal to zero, subtract 40 from each side:
40 - 40 = -12t2 + 14t + 40 - 40
which simplifies to
0 = -12t2 + 14t + 0
or
0 = -12t2 + 14t
Since the expression on the right is easily factored, factoring will be used in this solution. (The quadratic formula can also be used.)
Always start factoring with the greatest common factor (GCF). In this case the GCF is 2t.
0 = 2t (-6t + 7)
This is all the factoring that can be done.
In order for this product (multiplication) to be zero, one of the factors must be zero. So
2t = 0
or
-6t + 7 = 0
Solve each of these equations for t. Divide both sides of
2t = 0
by 2 and get
2t/2 = 0/2
which simplifies to:
t = 0
This says the ball was at height 40 at t = 0, the start (which we already knew).
For the other equation, subtract 7 from each side:
-6t + 7 - 7 = 0 - 7
giving
-6t + 0 = -7
or
-6t = -7
Now divide both sides of the equation by -6:
-6t/-6 = -7/-6
which simplifies to
t = 7/6
or, writing 7/6 as a mixed number:
t = 1 1/6
So the ball must return to a height of 40 after 1 1/6 seconds.