Q65P - University Physics with Modern Physics

Question

A 76.0 Kg rock is rolling horizontally at the top of a vertical cliff that is above the surface of a lake (Fig. P3.65). The top of the vertical face of a dam is located from the foot of the cliff, with the top of the dam level with the surface of the water in the lake. A level plain is 25m below the top of the dam. (a) What must be the minimum speed of the rock just as it leaves the cliff so that it will reach the plain without striking the dam? (b) How far from the foot of the dam does the rock hit the plain?

Step-by-Step Solution

Answer

(a) u =49.5m/s

 

(b) 50 m

Step 1: Introduction

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 $\mathrm{u} \sin \mathrm{\theta } \mathrm{and} \cos \mathrm{\theta }$ respectively.

Step 2: Given

Rock weight = 76Kg

 

Vertical cliff = 20m

 

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

                                                                                          

Level plain distance from dam =  25m

Step 3: (a) Minimum Speed of the rock

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,    

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

Step 4: (b) Distance from the foot of the dam where the rock hit the plain

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.