# Question d571f

Jul 1, 2015

The first drop travelled 1.0 m.

#### Explanation:

The idea behind this problem is that you need to determine how much time passes before the second drop of water falls from the tap.

You know that the first drop covers 0.25 m before the second drop is realeased. This means that it needed

$h = {\underbrace{{v}_{0}}}_{\textcolor{b l u e}{\text{=0}}} \cdot {t}_{1} + \frac{1}{2} \cdot g \cdot {t}_{1}^{2}$

$h = \frac{1}{2} \cdot g \cdot {t}_{1}^{2} \implies {t}_{1} = \sqrt{\frac{2 \cdot h}{g}}$

t_1 = sqrt((2 * 0.25cancel("m"))/(9.8cancel("m")/"s"^2)) = "0.226 s"

to cover 0.25 m.

The second drop is released 0.226 seconds after the first drop. Let's assume that the first drop travels a total time of $t$ seconds. This automatically means that the second rop will travel for a total time of $t - 0.226$, since it's realeased 0.226 seconds later than the first drop.

The difference between the distance covered by the first drop and the distance covered by the second drop must be 0.75 m, which means that you have

${h}_{1} - {h}_{2} = \text{0.75 m}$

underbrace(1/2 * g * t^2)_(color(blue)("first drop")) - underbrace(1/2 * g(t-0.226)^2)_(color(blue)("second drop")) = "0.75 m"

This is equivalent to

$\cancel{g \cdot {t}^{2}} - \cancel{g \cdot {t}^{2}} + 9.8 \cdot 0.452 \cdot t - 9.8 \cdot 0.0511 = 0.75 \cdot 2$

$4.43 \cdot t = 2.001 \implies t = \frac{2.001}{4.43} = \text{0.452 s}$

The distance between the two drops will be 0.75 m after the first drop travelled for 0.452 seconds.

Therefore,

${h}_{1} = \frac{1}{2} \cdot g \cdot {t}^{2}$

h_1 = 1/2 * 9.8 "m"/cancel("s"^2) * 0.452""^2cancel("s"^2) = color(green)("1.0 m")#