A body is thrown vertically upward from a point `A` 125 m above the ground. It goes up to a maximum height of 250 m above the ground and passes through `A` on its downward journey. The velocity of the body when it is at a height of 70 m above the ground is `(g=10 m//s^(2))`

Options

A. `20 m//s`
B. `50 m//s`
C. `60 m//s`
D. `80 m//s`

Explanation

To find the velocity of the body when it is at a height of 70 m above the ground, we can use the kinematic equation:

\[ v^2 = u^2 + 2g h \]

Where:
- \( v \) = final velocity
- \( u \) = initial velocity
- \( g \) = acceleration due to gravity (10 m/s²)
- \( h \) = height

### Step 1: Determine the initial conditions
The body is thrown from a height of 125 m and reaches a maximum height of 250 m. Therefore, when the body is at its maximum height, its initial velocity \( u \) at that point is 0 m/s.

### Step 2: Calculate the height difference
When the body is at a height of 70 m, we need to calculate the height difference from the maximum height (250 m) to 70 m. The height difference \( h \) is:

\[ h = 250 \, \text{m} - 70 \, \text{m} = 180 \, \text{m} \]

### Step 3: Use the kinematic equation
Now, we can substitute the values into the kinematic equation. Here, \( u = 0 \) (at maximum height), \( g = 10 \, \text{m/s}^2 \), and \( h = 180 \, \text{m} \):

\[ v^2 = 0 + 2 \times 10 \times 180 \]

### Step 4: Calculate \( v^2 \)
Calculating the right side:

\[ v^2 = 2 \times 10 \times 180 = 3600 \]

### Step 5: Find \( v \)
Now, take the square root to find \( v \):

\[ v = \sqrt{3600} = 60 \, \text{m/s} \]

### Conclusion
The velocity of the body when it is at a height of 70 m above the ground is **60 m/s**.

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