1. Initial setup

The ball is thrown from the roof of a building of height $10\,\text{m}$ with speed $10\,\text{m/s}$ at angle $30^\circ$ above the horizontal.

We need the distance of the ball from the throwing point when it is again at height $10\,\text{m}$ from the ground.

Since the throwing point itself is at height $10\,\text{m}$, this means we want the position of the ball when it returns to the same vertical level as the point of projection.

2. Resolve the initial velocity

Horizontal component: $u_x = 10\cos 30^\circ = 10\cdot \frac{\sqrt{3}}{2} = 5\sqrt{3}\,\text{m/s}$

Vertical component: $u_y = 10\sin 30^\circ = 10\cdot \frac{1}{2} = 5\,\text{m/s}$

3. Find the time when the ball is again at the same height

Vertical displacement from the throwing point is $y = u_y t - \frac{1}{2}gt^2$

At the same height as projection, $y=0$: $0 = 5t - \frac{1}{2}(10)t^2$ $0 = 5t - 5t^2$ $0 = 5t(1-t)$

So, $t=0 \quad \text{or} \quad t=1\,\text{s}$

Ignoring $t=0$ (instant of projection), the required time is $t=1\,\text{s}$

4. Find the horizontal distance in this time

Horizontal distance traveled: $x = u_x t = 5\sqrt{3}\times 1 = 5\sqrt{3}\,\text{m}$

Using $\sqrt{3}\approx 1.732$, $x = 5(1.732)=8.66\,\text{m}$

5. Distance from the throwing point

At this instant, the ball is at the same height as the throwing point, so the straight-line distance from the throwing point is just this horizontal distance: $\boxed{8.66\,\text{m}}$

6. Option check

• A: $5.20\,\text{m}$ — incorrect
• B: $4.33\,\text{m}$ — incorrect
• C: $2.60\,\text{m}$ — incorrect
• D: $8.66\,\text{m}$ — correct

Thus, the correct option is D.