Chapter 2, Problem 79GP

To determine

The initial velocity of the second stone thrown from a 4^th floor balcony

Answer to Problem 79GP

  $18.4\frac{\text{m}}{\text{s}}$

Explanation of Solution

Given:

Initial velocity of the first stone thrown $=v_{1o}=11\frac{\text{m}}{\text{s}}$

Height of the each floor balcony of a building $=\text{h}$

Height of the 12^th floor balcony of a building $=H_{12}=12\text{h}$

Initial position of the first stone $=y_{1o}=H_{12}=12\text{h}$

Final position of the first stone at the ground $=y_{1f}=0\text{m}$

Time of travel for first stone $=t_1=4.5\sec$

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

Initial velocity of the second stone thrown $=v_{2o}$

Height of the 4^th floor balcony of a building $=H_4=4\text{h}$

Initial position of the second stone $=y_{2o}=H_4=4\text{h}$

Final position of the second stone at the ground $=y_{2f}=0\text{m}$

Time of travel for second stone $=t_2=4.5\sec$

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

Formula Used:

Position of an object at any time in accelerated motion is given as

  $y_f=y_o+v_{o}t+(0.5)a{t}^2$

Calculation:

Final position of the first stone is given as

  $y_{1f}=y_{1o}+v_{1o}{t}_1+(0.5)a{t}_1{}^2$

Substituting the values,

  $\begin{array}{l}0=12h+(11)(4.5)+(0.5)(-9.8){(}^{4.5}\\ h=4.14\text{m}\end{array}$

Final position of the second stone is given as

  $y_{2f}=y_{2o}+v_{2o}{t}_2+(0.5)a{t}_2{}^2$

Substituting the values

  $\begin{array}{l}0=4h+v_{2o}(4.5)+(0.5)(-9.8){(}^{4.5}\\ 0=4(4.14)+v_{2o}(4.5)+(0.5)(-9.8){(}^{4.5}\\ v_{2o}=18.4\frac{\text{m}}{\text{s}}\end{array}$

Conclusion:

Therefore, the initial velocity of the second stone is $18.4\frac{\text{m}}{\text{s}}$