Question

A stone is dropped from the top of a building. When it crosses a point 5m below the top, another stone starts to fall from a point 25m below the top. Both stones reach the bottom of building simultaneously. The height of the building is

• A. 35m
• B. 45m
• C. 25m
• D. 50m

Answer 1

Answer: (B)

45m

Answer 2

Let the height of the building be $H$. Let the acceleration due to gravity be $g$.

Let the first stone be S1 and the second stone be S2.

S1 is dropped from the top of the building (initial position = 0, relative to the top) at time $t=0$. Its initial velocity $u_1 = 0$.

S2 is dropped from a point 25m below the top (initial position = 25m, relative to the top) at a later time $t_0$. Its initial velocity $u_2 = 0$.

First, let's find the time $t_0$ when S1 crosses the point 5m below the top.

For S1, the displacement is $s_1 = 5$m. Initial velocity $u_1 = 0$. Acceleration $a = g$.

Using the equation of motion $s = ut + \frac{1}{2}at^2$:

$5 = 0 \cdot t_0 + \frac{1}{2}gt_0^2$

$5 = \frac{1}{2}gt_0^2$

$t_0^2 = \frac{10}{g}$

$t_0 = \sqrt{\frac{10}{g}}$

According to the problem, the second stone S2 starts to fall at time $t_0$.

Both stones reach the bottom of the building simultaneously. Let $T$ be the total time taken by S1 to reach the bottom from the top.

The total displacement for S1 is $H$. Initial velocity $u_1 = 0$. Acceleration $a = g$.

Using the equation of motion $s = ut + \frac{1}{2}at^2$:

$H = 0 \cdot T + \frac{1}{2}gT^2$

$H = \frac{1}{2}gT^2$ (Equation 1)

The second stone S2 starts falling at time $t_0$ and reaches the bottom at time $T$. The duration of fall for S2 is $T - t_0$.

The starting point of S2 is 25m below the top. The total height of the building is $H$. So, the starting point of S2 is at a height $H-25$ from the bottom.

The displacement for S2 is $H-25$ (from its starting point 25m below the top to the bottom). Initial velocity $u_2 = 0$. Acceleration $a = g$.

Using the equation of motion $s = ut + \frac{1}{2}at^2$ for S2:

$H-25 = 0 \cdot (T - t_0) + \frac{1}{2}g(T - t_0)^2$

$H-25 = \frac{1}{2}g(T - t_0)^2$ (Equation 2)

Now substitute the value of $t_0 = \sqrt{\frac{10}{g}}$ into Equation 2:

$H-25 = \frac{1}{2}g\left(T - \sqrt{\frac{10}{g}}\right)^2$

Expand the square:

$H-25 = \frac{1}{2}g\left(T^2 - 2T\sqrt{\frac{10}{g}} + \left(\sqrt{\frac{10}{g}}\right)^2\right)$

$H-25 = \frac{1}{2}g\left(T^2 - 2T\sqrt{\frac{10}{g}} + \frac{10}{g}\right)$

Distribute $\frac{1}{2}g$:

$H-25 = \frac{1}{2}gT^2 - \frac{1}{2}g \cdot 2T\sqrt{\frac{10}{g}} + \frac{1}{2}g \cdot \frac{10}{g}$

$H-25 = \frac{1}{2}gT^2 - gT\sqrt{\frac{10}{g}} + 5$

From Equation 1, we know that $\frac{1}{2}gT^2 = H$. Substitute this into the equation above:

$H-25 = H - gT\sqrt{\frac{10}{g}} + 5$

Subtract $H$ from both sides:

$-25 = - gT\sqrt{\frac{10}{g}} + 5$

Subtract 5 from both sides:

$-30 = - gT\sqrt{\frac{10}{g}}$

Multiply by -1:

$30 = gT\sqrt{\frac{10}{g}}$

We can rewrite $g\sqrt{\frac{1}{g}}$ as $\sqrt{g^2 \cdot \frac{1}{g}} = \sqrt{g}$.

So, $gT\sqrt{\frac{10}{g}} = T\sqrt{g}\sqrt{10} = T\sqrt{10g}$.

$30 = T\sqrt{10g}$

Alternatively, $gT\sqrt{\frac{10}{g}} = gT \frac{\sqrt{10}}{\sqrt{g}} = T\sqrt{g}\sqrt{10}$.

$30 = T\sqrt{10g}$

Square both sides:

$30^2 = (T\sqrt{10g})^2$

$900 = T^2 (10g)$

$900 = 10 gT^2$

Divide by 10:

$90 = gT^2$

From Equation 1, $H = \frac{1}{2}gT^2$, which means $gT^2 = 2H$.

Substitute this into the equation $90 = gT^2$:

$90 = 2H$

$H = \frac{90}{2}$

$H = 45$ m.

The height of the building is 45m.