Story problem

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.