A boy playing on the roof of a 10m high building throws a ball with a speed of 10 m/s at an angle of 30° with the horizontal. How far from the throwing point will the ball be at the height of 10 m from the ground (g = 10 m/s², sin 30°=1/2, cos 30°=√3/2)

The correct option is 8.66 m.

We are given the following values from the problem statement:
- Height of the building = 10 m (the starting point of the ball)
- Speed of the ball, $u=10\mathrm{m/s}$
- Angle of projection, $\theta =30°$
- Acceleration due to gravity, $g=10{\mathrm{m/s}}^2$

The question asks for the distance of the ball from the throwing point when it reaches a height of 10 m from the ground. Since the ball is thrown from the roof of a 10 m high building, its launching height is 10 m from the ground. Therefore, the ball returns to the height of 10 m from the ground when its vertical displacement relative to the throwing point is zero ($y=0$).

First, let's resolve the initial velocity into horizontal and vertical components:
- Horizontal velocity component: $u_x=u\cos \theta$
- Vertical velocity component: $u_y=u\sin \theta$

To find the time $t$ when the ball returns to the launching level, we use the vertical equation of motion:
$y=u_{y}t-\frac{1}{2}g{t}^2$
Setting $y=0$:
$0=\left(u\sin \theta \right)t-\frac{1}{2}g{t}^2$

For $t\ne 0$, solving for $t$ gives the time of flight to reach the same level:
$t=\frac{2u\sin \theta }{g}$

Substituting the given values:
$t=\frac{2\times 10\times \sin 30°}{10}$
$t=2\times \frac{1}{2}=1s$

Next, we calculate the horizontal distance $x$ covered by the ball during this time interval:
$x=u_x\times t$
$x=\left(u\cos \theta \right)\times t$
$x=10\times \cos 30°\times 1$
$x=10\times \frac{\sqrt{3}}{2}=5\sqrt{3}m$

Using $\sqrt{3}\approx 1.732$:
$x=5\times 1.732=8.66m$

Since the vertical level of the ball at this point is the same as the throwing point, the straight-line distance from the throwing point is exactly equal to the horizontal distance, which is 8.66 m.