A key falls from a bridge that is 45 m above the water. It falls directly into a model boat, moving with constant velocity that is 12mfrom the point of impact when the key is released. What is the speed of the boat?

The height of the bridge is,$y=45\mathrm{m}$

The distance of the boat from point of impact is,$x=12\mathrm{m}$

Acceleration acting on the key is,$a=g=9.8\mathrm{m}/{\mathrm{s}}^2$

The time taken bythe key to fall through the given height is the same as the time taken by the boat to cover the given distance.By using the equation of kinematics,the speed of the boat can be determined.

The expression for the second equation of motion is,

$y=v_{0}t+\frac{1}{2}a{t}^2$ (i)

Here,$y$ is the vertical height, $v_0$is the initial speed, a is the acceleration and$t$ is the time.

Since the initial velocity of the keys is zero, therefore, equation (i) can be written as,

$$\begin{gathered} y=\left(0\right)t+\frac{1}{2}a{t}^2 \\ y=\frac{1}{2}a{t}^2 \end{gathered}$$

Substitute the values in the above equation.

$$\begin{gathered} 45m=\frac{1}{2}\times 9.8m/s^2\times t^2 \\ {t}^2=\frac{45m\times 2}{9.8m/s^2} \\ t=\sqrt{9.1837} \\ =3.03\mathrm{s} \end{gathered}$$

The time key takes to fall through height is the same as the time it takes the boat to cover the distance of 12 m to the point of impact.

The speed of the boat is calculated as,

$$\begin{gathered} v=\frac{x}{t} \\ =\frac{12\mathrm{m}}{3.03\mathrm{s}} \\ =3.96m/s \\ \approx 4.0\mathrm{m}/\mathrm{s} \end{gathered}$$

Thus, the speed of the boat is $4.0\mathrm{m}/\mathrm{s}$.