From the top of a cliff of height 150 m, a stone is thrown up with a velocity of `19.6 m//s`, at an angle of `60 ^(@)` with the vertical . Calculate the distance of the point where it strickes on the ground from the foot of the cliff .

Explanation

To solve the problem step by step, we will break down the motion of the stone into horizontal and vertical components, and then use kinematic equations to find the distance from the foot of the cliff where the stone strikes the ground.

### Step 1: Break down the initial velocity into components
The stone is thrown with an initial velocity \( u = 19.6 \, \text{m/s} \) at an angle of \( 60^\circ \) with the vertical. We need to find the horizontal and vertical components of this velocity.

- **Vertical component**:
\[
u_y = u \cos(60^\circ) = 19.6 \cos(60^\circ) = 19.6 \times 0.5 = 9.8 \, \text{m/s}
...