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**Hint**

The stone after reaching its highest point will take the same amount of time to pass from the point from where it was projected as the time it took to get to that point after being launched. Determine the time the stone takes to reach its highest point and then calculate the time taken when the rock free-falls to the ground under gravitational acceleration.

Formula used:

$\Rightarrow s = ut + \dfrac{1}{2}a{t^2}$ where $s$ is the distance covered by a stone when launched with an initial velocity $u$ in time $t$ under the influence of acceleration $a$.

$\Rightarrow v = u + at$ where $v$ is the final velocity of the object.

$\Rightarrow {v^2} = {u^2} + 2as$.

**Complete step by step answer**

When the stone is launched by the boy above the building, it will decelerate due to gravity and will eventually stop rising and start falling. The time taken by the stone to reach its highest point after which it starts falling can be calculated using the formula $v = u + at$where $u = 10\,m/s$, $v = 0\,m/s$ since the ball stops moving for a moment at its highest position and $a = - g = - 10m/{s^2}$ is the acceleration due to gravity. Hence,

$\Rightarrow 0 = 10 - 10t$

$\Rightarrow t = 1s$

The ball reaches its highest point in 1s and will take the same amount to come back down to the point from where it was projected. Hence, it will pass the point from where it was projected in 2s.

The height reached by the stone at its topmost point can be calculated from

$\Rightarrow s = ut + \dfrac{1}{2}a{t^2}$

On substituting $u = 10m/s$ and $a = - g = - 10m/{s^2}$, we can calculate the height obtained by the stone as:

$\Rightarrow s = 10 + \dfrac{1}{2}( - 10) $

$\Rightarrow 5m $

Hence the stone will rise to a height of \[40 + 5{\text{ }} = {\text{ }}45{\text{ }}m\] above the ground. The time the ball will take to fall to the ground can be calculated again from

$\Rightarrow s = ut + \dfrac{1}{2}a{t^2}$

where $u = 0m/s$, and the downward distance to be covered $s = - 45m$

$\Rightarrow - 45 = \dfrac{1}{2}( - 10){t^2}$

$\Rightarrow {t^2} = 9$

Taking the square root on both sides, we get

$\Rightarrow t = 3s$

Hence it will strike the ground $3 + 1 = 4s$ after being launched. The velocity with which the ball strikes the ground after falling from the highest point can be calculated from

$\Rightarrow {v^2} = {u^2} + 2as$ where we substitute $u = 0m/s$ and $s = 45m$ get

$\Rightarrow {v^2} = 900$

$\therefore v = 30m/s$ but since the velocity is pointing in the downwards direction, $v = - 30m/s$

**Hence the correct option is option (B).**

**Note**

To answer this type of question, the key is to know the phenomena of a body thrown vertically upwards under the influence of gravity, its trajectory, and its properties while it’s rising and falling. We must also be careful about the sign of different quantities such as distance, speed and acceleration which are generally taken to be positive in the upward direction and negative in the downward direction

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