91Ó°ÊÓ

The velocity contains two components, one is a horizontal component and another one is vertical.

According to the newton’s laws of motion,

$s=ut+\frac{1}{2}a{t}^2$

Where, s, u, t, a, are displacement, initial velocity, time, and acceleration.

For the vertical and horizontal motion, the velocity component will be$\mathrm{u}\sin \mathrm{θ}\mathrm{and}\cos \mathrm{θ}$ respectively.

Rock weight = 76Kg

Vertical cliff = 20m

The top of the vertical face of the dam = 100m

Level plain distance from dam = 25m

Initial velocity is zero, and gravity for horizontal motion is zero.

For the vertical motion,

$$\begin{gathered} \mathrm{s}={\mathrm{gt}}^2 \\ 20\mathrm{m}=\left(9.8\mathrm{m}/{\mathrm{s}}^2\right){\mathrm{t}}^2 \\ \mathrm{t}=2.02\mathrm{s} \end{gathered}$$

For the horizontal motion,

$$\begin{gathered} \mathrm{s}=\mathrm{ut} \\ \mathrm{u}=\frac{100\mathrm{m}}{2.02\mathrm{s}} \\ \mathrm{u}=49.5\mathrm{m}/\mathrm{s} \end{gathered}$$

For the vertical motion,

$$\begin{gathered} \mathrm{s}={\mathrm{at}}^2 \\ 45\mathrm{m}=\left(9.8\mathrm{m}/{\mathrm{s}}^2\right){\mathrm{t}}^2 \\ \mathrm{t}=3.03\mathrm{s} \end{gathered}$$

For the horizontal motion,

$$\begin{gathered} \mathrm{s}=\mathrm{ut} \\ \mathrm{s}=\left(49.5\mathrm{m}/\mathrm{s}\right)\left(3.03\mathrm{s}\right) \\ \mathrm{s}=150\mathrm{m} \end{gathered}$$

$150\mathrm{m}-100\mathrm{m}=50\mathrm{m}$

The rock land 50m beyond the foot of the dam.