Chapter 11, Problem 96A

To determine

To compute: - The rank of the five-clay potsaccording to their speed when they strike to the ground from least to greatest.

Answer to Problem 96A

  $v_D<v_A=v_E<v_B=v_C$

Explanation of Solution

Given info:

Height from the ground, $h=5\text{ m}$

Acceleration due to gravity, $g=9.8{\text{ m/s}}^2$

(A)1 kg dropped from rest with initial velocity, $u_A=0$

(B)1 kg thrown downward with initial velocity, $u_B=2\text{ m/s}$

(C)1 kg thrown upward with initial velocity, $u_C=2\text{ m/s}$

(D)1 kg thrown horizontally with initial velocity, $u_D=2\text{ m/s}$

(E)2 kg dropped from rest, $u_E=0$

Formula used:

For motion in straight line with constant acceleration $a$ :

Mathematically it can be represented as,

  $v^2=u^2+2ax$

Where,

  $x=$ Displacement.

  $u=$ Initial velocity.

  $v=$ Final velocity.

  $a=$ Constant acceleration.

Calculation:

Case A:

Clay is moving downward, therefore acceleration:

  $a=g=9.8{\text{ m/s}}^2$

Using the formula of kinematics equation:

  $\begin{array}{c}{v}_A{}^2=u_A{}^2+2gh\\ =0^2+2\times 9.81\times 5\\ v_A=9.89\text{ m/s}\end{array}$

Case (B):

Clay pot is moving downward, therefore acceleration:

  $a=g=9.8{\text{ m/s}}^2$

Using the formula of kinematics equation:

  $\begin{array}{c}{v}_B{}^2=u_B{}^2+2gh\\ =2^2+2\times 9.81\times 5\\ v_B=10.09\text{ m/s}\end{array}$

Case (C):

Let height reached by clay pot above the rooftop be $H_C$

Therefore, formula of maximum height reached by projectile:

  $H_C=\frac{u_C{}^2{\sin }^2\theta }{2g}$

Plugging in the given values:

  $\begin{array}{c}{H}_C=\frac{{\left(\text{2}\right)}^{\text{2}}{\text{sin}}^{\text{2}}\left({\text{90}}^{\text{0}}\right)}{\text{2}\left(\text{9}\text{.8}\right)}\\ {\text{H}}_C\text{=}\text{0}\text{.204}\text{m}\end{array}$

Total distance covered by pot from height $H_C$ to ground:

  $\begin{array}{c}x=5+0.204\\ =5.204\text{ m}\end{array}$

Using the formula of kinematics equation:

  $\begin{array}{c}{v}_C{}^2=u_C{}^2+2gx\\ =0^2+2\times 9.81\times 5.204\\ v_C=10.099\text{ m/s}\end{array}$

Case (D):

Since, pot thrown horizontally, therefore: $\theta =0^o$

Maximum horizontal distance covered by pot is the range of pot,

  $\begin{array}{c}x=R=\frac{u^2\sin 2\theta }{g}\\ =0\end{array}$

Using the formula of kinematics equation:

  $\begin{array}{c}{v}_D{}^2=u_D{}^2+2gx\\ =2^2+2\times 9.81\times 0\\ v_D=2\text{ m/s}\end{array}$

Case (E):

Using the formula of kinematics equation:

  $\begin{array}{c}{v}_E{}^2=u_E{}^2+2gh\\ =0^2+2\times 9.81\times 5\\ v_E=9.89\text{ m/s}\end{array}$

Conclusion:

The rank of the speed:

  $v_D<v_A=v_E<v_B=v_C$