From the top of a building , a ball is thrown horizontally. The ball strickes the ground after 5 s at an angle of `60^(@)` to the horizontal. Calculate the speed with which the ball was projected .

Answer

To solve the problem of a ball thrown horizontally from the top of a building and striking the ground at an angle of 60 degrees after 5 seconds, we can follow these steps:

### Step 1: Understand the motion components
The ball is thrown horizontally, which means its initial vertical velocity is 0. The motion can be analyzed in two dimensions: horizontal (x-direction) and vertical (y-direction).

### Step 2: Identify the vertical motion
In the vertical direction, the ball is influenced by gravity. The final vertical velocity \( v_y \) can be expressed using the kinematic equation:
\[ v_y = u_y + a \cdot t \]
Where:
- \( u_y = 0 \) (initial vertical velocity)
- \( a = g = 10 \, \text{m/s}^2 \) (acceleration due to gravity)
- \( t = 5 \, \text{s} \) (time of flight)

Substituting the values:
\[ v_y = 0 + 10 \cdot 5 = 50 \, \text{m/s} \]

### Step 3: Relate the final velocity to the angle
The ball strikes the ground at an angle of 60 degrees to the horizontal. We can relate the vertical and horizontal components of the final velocity using trigonometric functions:
- \( v_y = v \sin(60^\circ) \)
- \( v_x = v \cos(60^\circ) \)

Where \( v \) is the magnitude of the final velocity. From the previous step, we have:
\[ v_y = 50 \, \text{m/s} \]
Thus:
\[ 50 = v \sin(60^\circ) \]
Using \( \sin(60^\circ) = \frac{\sqrt{3}}{2} \):
\[ 50 = v \cdot \frac{\sqrt{3}}{2} \]
Solving for \( v \):
\[ v = \frac{50 \cdot 2}{\sqrt{3}} = \frac{100}{\sqrt{3}} \, \text{m/s} \]

### Step 4: Calculate the horizontal velocity
The horizontal component of the velocity is given by:
\[ u = v \cos(60^\circ) \]
Using \( \cos(60^\circ) = \frac{1}{2} \):
\[ u = \frac{100}{\sqrt{3}} \cdot \frac{1}{2} = \frac{50}{\sqrt{3}} \, \text{m/s} \]

### Final Answer
The speed with which the ball was projected is:
\[ u = \frac{50}{\sqrt{3}} \, \text{m/s} \]

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