Chapter 2, Problem 78GP

(a)

To determine

Time required by first stone to reach the ground.

Answer to Problem 78GP

Time required by first stone to reach the ground is 5.63 s

Explanation of Solution

Given:

Initial velocity of second stone which is thrown after 2 seconds is 25 m/s.

Both the stones land at the same time.

Formula used:

  $\begin{array}{l}S=ut+\frac{1}{2}a{t}^2\\ v^2=u^2+2gS\end{array}$

Calculation:

Here,

  $\begin{array}{l}{t}_2=t_1-2\\ u=25\text{ m/s}\\ g=\text{9}{\text{.8 m/s}}^2\end{array}$

t₁ is the time taken by the first stone and t₂ is the time taken by the second stone.

Using kinematical equation,

  $S=ut+\frac{1}{2}a{t}^2$

For first stone,

  $S_1=\frac{1}{2}g{t}_1^2(1)$

For second stone,

  $S_2=\frac{1}{2}g{t}_2^2$

But $t_2=t_1-2$

Hence,

  $S_2=25\times \left(t_1-2\right)+\frac{1}{2}\times g\times {\left(t_1-2\right)}^2(2)$

To find the value of t₁, divide both equations.

  $\frac{S_1}{S_2}=\frac{\frac{1}{2}g{t}_1^2}{25\times (t-2)+\frac{1}{2}\times g\times {\left(t_1-2\right)}^2}$

But both stones are falling from same height. So

  $S_1=S_2$

  $\begin{array}{l}\frac{1}{2}g{t}_1^2=25\times (t_1-2)+\frac{1}{2}\times g\times {\left(t_1-2\right)}^2\\ t_1=5.63\sec \end{array}$

Conclusion:

Time required by first stone to reach the ground is 5.63 s

(b)

To determine

Height of the building

Answer to Problem 78GP

Height of the building is 155.3 m.

Explanation of Solution

Given:

Initial velocity of second stone which is thrown after 2 seconds is 25 m/s.

Both the stones land at the same time.

Formula used:

  $\begin{array}{l}S=ut+\frac{1}{2}a{t}^2\\ v^2=u^2+2gS\end{array}$

Calculation:

To find the height of the building,

  $\begin{array}{l}{S}_1=\frac{1}{2}g{t}_1^2\\ S_1=\frac{1}{2}\times 9.8\times {\left(5.63\right)}^2\\ S_1=155.3\text{ m}\end{array}$

Conclusion:

Height of the building is 155.3 m.

(c)

To determine

Speeds of both the stones.

Answer to Problem 78GP

Speeds of both the stones 55 m/s.

Explanation of Solution

Given:

Initial velocity of second stone which is thrown after 2 seconds is 25 m/s.

Both the stones land at the same time.

Formula used:

  $\begin{array}{l}S=ut+\frac{1}{2}a{t}^2\\ v^2=u^2+2gS\end{array}$

Calculation:

  $\begin{array}{l}{S}_1=S_2=155.3\text{m}\\ g=9.8m/s^2\end{array}$

As height of building is same, hence velocity before reaching the ground would be,

  $\begin{array}{l}{V}^2=2g{S}_1=2g{S}_2=2\times 9.8\times 155.3\\ V=55\text{ m/s}\end{array}$

Conclusion:

Speeds of both the stones will be 55 m/s.