A stone is thrown horizontally with an initial speed of $10\ \mathrm{\displaystyle \frac{m}{s}}$ from a bridge. Assuming that air resistance is negligible, how long would it take the stone to strike the water $80\ \mathrm{m}$ below the bridge?
A) $1\ \mathrm{s}$
B) $2\ \mathrm{s}$
C) $4\ \mathrm{s}$
D) $8\ \mathrm{s}$
A) $1\ \mathrm{s}$
B) $2\ \mathrm{s}$
C) $4\ \mathrm{s}$
D) $8\ \mathrm{s}$
Apply Big Five #3 to the vertical motion, calling down the positive direction:
$$\Delta y=v_{0y}t+\frac{1}{2}a_yt^2$$
$$\Delta y=\frac{1}{2}a_yt^2$$
$$\Delta y=\frac{1}{2}gt^2$$
$$t=\sqrt{\frac{2\Delta y}{g}}$$
$$t=\sqrt{\frac{2(80\ \mathrm{m})}{10\ \mathrm{\displaystyle \frac{m}{\ s^2}}}}$$
$$t=4\ \mathrm{s}$$
Note that the stone’s initial horizontal speed $\left(v_{0x}=10\ \mathrm{\displaystyle \frac{m}{s}}\right)$ is irrelevant.