A stone is dropped from the roof of a high building. A second stone is dropped 1.30 s later. How far apart are the stones when the second one has reached a speed of $\mathbf{12}\mathbf{.}\mathbf{0}\mathbf{m}\mathbf{/}\mathbf{s}$.

The first stone is dropped from the roof of the building, and it falls under the effect of gravitational force.

As a result, its acceleration is equal to the acceleration due to gravity.

The second stone is dropped sometime after the first one is dropped. It also moves with the same acceleration under the effect of gravity.

The first equation of motion is.

$v=u+at\dots \left(i\right)$

Here, vis the final velocity of the stone, u is the initial velocity of the stone, a is the acceleration of the stone, t is the time taken by the stone.

• The initial speed of the second stone,$u_2=0\mathrm{m}/\mathrm{s}$
• Acceleration of the second stone,$a=g=9.81\mathrm{m}/{\mathrm{s}}^2$
• The final speed of the second stone,$v_2=12\mathrm{m}/\mathrm{s}$

Using the first equation of motion, you get the final velocity of the stone from equation (i) as.

$\begin{array}{c}{v}_2=u_2+a{t}_2\\ =u_2+g{t}_2\end{array}$

Substitute the values as 0 m/s for $u_2$, 12 m/s for $v_2$, and $9.81\mathrm{m}/{\mathrm{s}}^2$ for g.

$\begin{array}{c}12\mathrm{m}/\mathrm{s}=0\mathrm{m}/\mathrm{s}+9.81\mathrm{m}/{\mathrm{s}}^2\times t_2\\ t_2=1.22\mathrm{s}\end{array}$

The second stone is dropped after 1.30 s, equal to the extra time taken by the first stone. The total time taken by the first stone is the sum of time taken by the second stone to reach the ground and the extra time.

The total time taken by the first stone is.

$t_1=t+t_2$

Here, t is the extra time taken by the first stone at $1.30\mathrm{s}$and $t_2$is the time taken by the second stone.

$\begin{array}{c}{t}_1=1.30\mathrm{s}+1.22\mathrm{s}\\ =2.52\mathrm{s}\end{array}$

The distance moved by the first stone can be expressed using the second equation of motion.

$\begin{array}{c}{s}_1=u_1{t}_1+\frac{1}{2}a{t}_1^2\\ s_1=u_1{t}_1+\frac{1}{2}g{t}_1^2\end{array}$

Here,$u_1$is the initial speed of the first stone at 0 m/s.

Substitute the values as 0 m/s for $u_1$$2.52\mathrm{s}$, for $t_1$, and $9.81\mathrm{m}/{\mathrm{s}}^2$ for g in the above equation.

role="math" localid="1643106868189" $\begin{array}{c}{s}_1=0\mathrm{m}/\mathrm{s}\times t_1+\frac{1}{2}\times 9.81\mathrm{m}/{\mathrm{s}}^2\times {\left(2.52\mathrm{s}\right)}^2\\ =31.149\mathrm{m}\end{array}$

The distance traveled by the second stone is.

$\begin{array}{c}{s}_2=u_2{t}_2+\frac{1}{2}a{t}_2^2\\ s_2=u_2{t}_2+\frac{1}{2}g{t}_2^2\end{array}$

Substitute the values 0 m/s for $u_2$, $1.22\mathrm{s}$ for$t_2$ and $9.81\mathrm{m}/{\mathrm{s}}^2$ for g in the above equation.

$\begin{array}{c}{s}_2=0\mathrm{m}/\mathrm{s}\times t_2+\frac{1}{2}\times 9.81\mathrm{m}/{\mathrm{s}}^2\times {\left(1.22\mathrm{s}\right)}^2\\ =7.3\mathrm{m}\end{array}$

The distance between the first and the second stone is.

$s=s_1-s_2$

Substitute the values 7.3 m for $s_2$and $31.149\mathrm{m}$for $s_1$in the above equation.

$\begin{array}{c}s=s_1-s_2\\ =31.149\mathrm{m}-7.3\mathrm{m}\\ =23.849\mathrm{m}\end{array}$

Thus, the distance between the two stones is $23.849\mathrm{m}$.