# A pelican ﬂying at 7.6 m/s along a horizontal path drops a ﬁsh from a height of 2.7 m above the water. How far does the ﬁsh travel horizontally before hitting the water below?

Nov 9, 2015

$5.64 \text{m}$

#### Explanation:

Get the time of flight first by considering the vertical component of motion:

$s = \frac{1}{2} \text{g} {t}^{2}$

$\therefore t = \sqrt{\frac{2 \times 2.7}{9.8}}$

$t = 0.742 \text{s}$

Now we consider the horizontal component of the motion:

$s = v \times t = 7.6 \times 0.742 = 5.64 \text{m}$