A ball is thrown horizontally from a cliff such that it strikes the ground after `5s`.The line of sight makes an angle `37^(@)` with the horizontal.The initial velocity of projection in `ms^(-1)` is

Answer

To solve the problem step by step, we will analyze the motion of the ball thrown horizontally from the cliff and use the given information to find the initial velocity of projection.

### Step 1: Understand the motion
The ball is thrown horizontally from a height (cliff) and strikes the ground after 5 seconds. The angle of the line of sight to the horizontal when the ball strikes the ground is 37 degrees.

### Step 2: Identify the height of the cliff
Using the time of flight (t = 5 s), we can find the height (H) of the cliff using the formula for vertical motion under gravity:
\[ H = \frac{1}{2} g t^2 \]
Where \( g \) is the acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \)).
Substituting the values:
\[ H = \frac{1}{2} \cdot 10 \cdot (5)^2 = \frac{1}{2} \cdot 10 \cdot 25 = 125 \, \text{m} \]

### Step 3: Relate height and range using the angle
The angle of the line of sight (37 degrees) gives us a relationship between the height (H) and the horizontal distance (R) traveled by the ball. Using the tangent function:
\[ \tan(37^\circ) = \frac{H}{R} \]
From trigonometric tables, we know:
\[ \tan(37^\circ) \approx \frac{3}{4} \]
Thus:
\[ \frac{3}{4} = \frac{125}{R} \]
From this, we can solve for R:
\[ R = \frac{125 \cdot 4}{3} = \frac{500}{3} \approx 166.67 \, \text{m} \]

### Step 4: Find the initial horizontal velocity (u)
The horizontal distance (R) can also be expressed in terms of the initial horizontal velocity (u) and time (t):
\[ R = u \cdot t \]
Substituting the known values:
\[ \frac{500}{3} = u \cdot 5 \]
Now, solving for u:
\[ u = \frac{500}{3 \cdot 5} = \frac{500}{15} = \frac{100}{3} \approx 33.33 \, \text{m/s} \]

### Step 5: Conclusion
The initial velocity of projection (u) is approximately:
\[ u \approx 33.33 \, \text{m/s} \]

---