A stone is projected from the ground with a velocity of 20 m/s at angle `30^(@)` with the horizontal. After one second it clears a wall then find height of the wall. `(g=10 ms^(-2))`

Answer

To find the height of the wall that the stone clears after being projected, we can follow these steps:

### Step 1: Resolve the initial velocity into components
The stone is projected with an initial velocity \( u = 20 \, \text{m/s} \) at an angle \( \theta = 30^\circ \) with the horizontal. We need to find the vertical component of the initial velocity, \( u_y \).

\[ u_y = u \sin \theta = 20 \sin 30^\circ \]

Since \( \sin 30^\circ = \frac{1}{2} \):

\[ u_y = 20 \times \frac{1}{2} = 10 \, \text{m/s} \]

### Step 2: Use the equation of motion to find the vertical displacement
We need to find the vertical displacement \( s_y \) after \( t = 1 \, \text{s} \). The equation of motion we will use is:

\[ s_y = u_y t + \frac{1}{2} a t^2 \]

Here, \( a = -g \) (acceleration due to gravity, acting downwards), so \( g = 10 \, \text{m/s}^2 \).

Substituting the values:

\[ s_y = 10 \times 1 + \frac{1}{2} \times (-10) \times (1)^2 \]

### Step 3: Calculate the vertical displacement
Now we can calculate \( s_y \):

\[ s_y = 10 - \frac{1}{2} \times 10 \times 1 = 10 - 5 = 5 \, \text{m} \]

### Step 4: Conclusion
The height of the wall \( h \) that the stone clears after 1 second is equal to the vertical displacement \( s_y \):

\[ h = s_y = 5 \, \text{m} \]

Thus, the height of the wall is **5 meters**.

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