Chapter 2, Problem 69GP

To determine

Distance between the two stones

Answer to Problem 69GP

29 m

Explanation of Solution

Given:

For the second stone:

Initial velocity of the second stone $=v_{2o}=0\frac{\text{m}}{\text{s}}$

Final velocity of the second stone $=v_{2f}=12\frac{\text{m}}{\text{s}}$

Acceleration of the second stone $=a_2=9.8\frac{\text{m}}{{\text{s}}^{\text{2}}}$

Time of travel for the second stone $=t_2=t$

Displacement of the second stone $=x_2$

For the first stone:

Time of travel for the first stone $=t_1=t_2+1.50$

Displacement of the first stone $=x_1$

Initial velocity of the first stone $=v_{1o}=0\frac{\text{m}}{\text{s}}$

Acceleration of the first stone $=a_1=9.8\frac{\text{m}}{{\text{s}}^{\text{2}}}$

Distance between the two stones $=d$

Formula Used:

Acceleration is given as

  $a=\frac{v_f-v_o}{t}$

Displacement is given as

  $x=v_{o}t+(0.5)a{t}^2$

Calculation:

Consider the motion of the second stone while falling down.

Average acceleration of the second stone is given as

  $a_2=\frac{v_{2f}-v_{2o}}{t}$

Inserting the values

  $\begin{array}{l}9.8=\frac{12-0}{t}\\ t=1.22\sec \end{array}$

Displacement of the second stone is given as

  $\begin{array}{l}{x}_2=v_{2o}{t}_2+(0.5)a_2{t}_2{}^2\\ x_2=(0)t+(0.5)(9.8)t^2\\ x_2=(0)(1.22)+(0.5)(9.8){(}^{1.22}\\ x_2=7.3\text{m}\end{array}$

Displacement of the first stone is given as

  $\begin{array}{l}{x}_1=v_{1o}{t}_1+(0.5)a_1{t}_1{}^2\\ x_1=v_{1o}(t_2+1.5)+(0.5)a_2{(}^{t_2}\\ x_1=(0)(1.22+1.5)+(0.5)(9.8){(}^{1.22}\\ x_1=36.3\text{m}\end{array}$

Distance between the two stones is given as

  $\begin{array}{l}d=x_1-x_2\\ d=36.3-7.3\\ d=29\text{m}\end{array}$

Conclusion:

Hence the distance between the two stones is 29 m.