Chapter 6.1, Problem 6PP

To determine

The velocity of the rock when it hits the ground.

Answer to Problem 6PP

The velocity of the rock while hitting the ground is $v=32.0\text{m/s}$ at an angle $\theta =82.4°$ .

Explanation of Solution

Given info:

A rock is thrown from the top of cliff.

The height of the cliff is $y=50.0\text{m}$ .

The initial velocity of the of the stone is $v_{\text{i}}=7.0\text{m/s}$ .

The angle of inclination of throw is $\theta =53.0°$ .

Formula used:

The free body diagram of throwing of stone from the top of the cliff as shown below:

Refer the Figure.

Consider the horizontal and vertical component of initial velocity are $v_{\text{x}}$ and $v_{\text{y}}$ .

Find the horizontal component of the initial velocity as follows:

$v_{\text{x}}=v_{\text{i}}\cos \theta$

Find the vertical component of the initial velocity as follows:

Apply Equation of kinematics,

$\begin{array}{l}{v}_{\text{y}}^2=v_{\text{iy}}^2+2gy\\ v_{\text{y}}^2={\left(v_{\text{i}}\sin \theta \right)}^2+2gy\end{array}$

Find the magnitude of the velocity as follows:

$\begin{array}{l}v=\sqrt{v_{\text{x}}^2+v_{\text{y}}^2}\\ v=\sqrt{{\left(v_{\text{i}}\cos \theta \right)}^2+\left\{{\left(v_{\text{i}}\sin \theta \right)}^2+2gy\right\}}\\ v=\sqrt{\left(v_{\text{i}}^2{\cos }^2\theta +v_{\text{i}}^2{\sin }^2\theta \right)+2gy}\\ v=\sqrt{v_{\text{i}}^2+2gy}\end{array}$

Calculation:

Consider the value of the acceleration due to gravity as $g=9.80{\text{m/s}}^2$ .

$\begin{array}{l}v=\sqrt{7^2+2\times 9.80\times 50}\\ v=\sqrt{7^2+2\times 9.80\times 50}\\ v=32.0\text{m/s}\end{array}$

Find the direction of the velocity $v$ as follows:

  $\begin{array}{l}\tan \theta =\left(\frac{v_y}{v_x}\right)\\ \tan \theta =\left(\frac{\sqrt{v_{\text{iy}}^2+2gy}}{v_{\text{i}}\cos \theta }\right)\\ \tan \theta =\left(\frac{\sqrt{v_{\text{i}}^2{\sin }^2\theta +2gy}}{v_{\text{i}}\cos \theta }\right)\end{array}$

  $\begin{array}{l}\tan \theta =\left(\frac{\sqrt{7^2{\sin }^2\left(53°\right)+2\times 9.80\times 50}}{7\times \cos \left(53°\right)}\right)\\ \tan \theta =\left(\frac{\sqrt{7^2{\sin }^2\left(53°\right)+2\times 9.80\times 50}}{7\times \cos \left(53°\right)}\right)\\ \theta ={\tan }^{-1}\left(\frac{31.8}{4.2}\right)\\ \theta =82.4°\end{array}$

Conclusion:

Thus, the velocity of the rock while hitting the ground is $v=32.0\text{m/s}$ at an angle $\theta =82.4°$ .