# Story problem

• Jun 7th 2007, 05:33 PM
Eponine
Story problem
You are standing on a bridge over a creek, holding a stone 20 ft. above the water
You release the stone. How long will it take the stone to hit the water?
You take another stone and toss it straight up with an initial velocity of 30 ft. per second. How long will it take the stone to hit the water?
If you throw a stone straight up in the air with the initial velocity of 50 ft. per second, could the stone reach a hight of 60 ft. above the water?

Can you help me? These story problemas always confuse me.~Eponine
• Jun 7th 2007, 08:22 PM
Soroban
Hello, Eponine!

You are expected to know the "free fall formula":
. . $y \;=\;h_o + v_ot - 16t^2$

where: . $y$ = height of the object, $h_o$ = initial height of the object,
$v_o$ = initial velocity of the object.

Quote:

You are standing on a bridge over a creek, holding a stone 20 ft. above the water

(a) You release the stone. .How long will it take the stone to hit the water?

We have: . $h_o = 20,\;v_o = 0$
The equation is: . $y \;=\; 20 - 16t^2$

"Hit the water" means the height of the stone is zero.
When is $y = 0$?
. . $20 - 16t^2\:=\:0\quad\Rightarrow\quad t^2 \:=\:\frac{20}{16}$

Therefore, the stone hits the water at: $t \,=\,\frac{\sqrt{5}}{2}$ seconds.

Quote:

You take another stone and toss it straight up with an initial velocity of 30 ft/sec.
How long will it take the stone to hit the water?

We have: . $h_o = 20,\;v_o = 30$
The equation is: . $y \;=\;20 + 30t - 16t^2$

When is $y = 0$ ?
. . $20 + 30t - 16t^2\:=\:0\quad\Rightarrow\quad 8t^2 - 15t - 10 \:=\:0$

Quadratic Formula: . $t \;=\;\frac{-(\text{-}15) \pm \sqrt{(\text{-}15)^2 - 4(8)(\text{-}10)}}{2(8)} \;=\;\frac{15\pm\sqrt{545}}{16}$

Therefore, the stone hits the water at: . $t \;=\;\frac{15 + \sqrt{545}}{16} \:\approx\:2.4$ seconds.

Quote:

If you throw a stone straight up in the air with the initial velocity of 50 ft/sec,
could the stone reach a height of 60 ft. above the water?

The equation is: . $y \;=\;20 + 50t - 16t^2$

There are at least two ways to answer this question. .(The answer is NO.)

[1] When is $y = 60$ ?

. . . $20 + 50t - 16t^1\:=\:60\quad\Rightarrow\quad 8t^2 - 25t + 20\:=\:0$

Quadratic Formula: . $t \;=\;\frac{-(\text{-}25) \pm \sqrt{(\text{-}25)^2 - 4(8)(20)}}{2(8)} \;=\;\frac{25 \pm\sqrt{-15}}{16}$ no real roots

Therefore, the stone never reaches a height of 60 feet.

[2] The graph of $y \:=\:20 +50t - 16t^2$ is a down-opening parabola.
. . .Its maximum point occurs at its vertex.
The vertex occurs at: . $t \:=\:\frac{-b}{2a}$

We have: . $a = -16,\;b = 50,\;c = 20$
The vertex is: . $t \:=\:\frac{-50}{2(-16)} \:=\:\frac{25}{16}$

When $x = \frac{25}{16},\;y \;=\;20 + 50\left(\frac{25}{16}\right) - 16\left(\frac{25}{16}\right)^2 \;=\;59.0625$ feet.

The maximum height is only $59\frac{1}{16}$ feet.
. . It never reaches 60 feet.