SOLUTION: An object that is projected straight downward with initial velocity of v feet per second is at a height from ground {{{ h=-16t^2-vt+s }}} , where s = initial height in feet, and t

Question 193178: An object that is projected straight downward with initial velocity of v feet per second is at a height from ground , where s = initial height in feet, and t = time in seconds. If Bianca is standing on a ledge 46.75 feet above the ground and throws a penny straight down with an initial velocity of 12 feet per second, in how many seconds will it reach the ground? -- I tried as hard as I can to solve this word problem but the answer was not reasonable... please help me, thank you!
Found 2 solutions by nerdybill, josmiceli:
Answer by nerdybill(7384) (Show Source): You can put this solution on YOUR website!
An object that is projected straight downward with initial velocity of v feet per second is at a height from ground , where s = initial height in feet, and t = time in seconds. If Bianca is standing on a ledge 46.75 feet above the ground and throws a penny straight down with an initial velocity of 12 feet per second, in how many seconds will it reach the ground?
.

The problem gives you:
s (initial height) as 46.75 feet
v (initial velocity) as 12 feet/sec
.
Since they want to know when it will reach the ground -- h (height) would be 0.
So, set h = 0 and solve for 't':
.

Since we can't factor, use the quadratic equation to solve. Doing so yields:
t={-2.125, 1.375}
Throw out the negative solution -- this leaves:
t = 1.375 seconds
.
Details of quadratic follows:

Solved by pluggable solver: SOLVE quadratic equation with variable

Quadratic equation (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=3136 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: -2.125, 1.375. Here's your graph:

Answer by josmiceli(19441) (Show Source): You can put this solution on YOUR website!
I think the key is that is the height from the ground and the
problem wants to know when the penny reaches ground, or when

given:
ft/sec
ft
Another key is that signs are already figured into the
equation, so I don't have to worry about signs
--------------------

Using the quadratic formula:

If the numerator is (+), then is negative, and I can't have
negative time, so the numerator must be negative

sec
The penny will reach the ground in 1.375 seconds
check answer:

I don't have a calculator, but I think this checks out
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