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. `50 ms^(-1)`
B. `60 ms^(-1)`
C. `80 ms^(-1)`
D. `20 ms^(-1)`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the problem
A body is thrown vertically upward from point A, which is 125 m above the ground. It reaches a maximum height of 250 m above the ground. We need to find the velocity of the body when it is at a height of 70 m above the ground.

### Step 2: Determine the initial velocity (u)
To find the initial velocity (u) of the body when it is thrown from point A, we can use the kinematic equation:
\[ v^2 = u^2 - 2gh \]
At the maximum height, the final velocity (v) is 0 m/s, and the height (h) is the distance from point A to the maximum height:
\[ h = 250 \, \text{m} - 125 \, \text{m} = 125 \, \text{m} \]

Substituting the values into the equation:
\[ 0 = u^2 - 2 \cdot 10 \cdot 125 \]
\[ u^2 = 2 \cdot 10 \cdot 125 \]
\[ u^2 = 2500 \]
\[ u = \sqrt{2500} = 50 \, \text{m/s} \]

### Step 3: Calculate the velocity at 70 m height
Now, we need to find the velocity (v) of the body when it is at a height of 70 m above the ground. The height from point A to the height of 70 m is:
\[ h' = 125 \, \text{m} - 70 \, \text{m} = 55 \, \text{m} \]

We will use the kinematic equation again:
\[ v^2 = u^2 - 2gh' \]
Here, \( u = 50 \, \text{m/s} \) and \( h' = 55 \, \text{m} \):
\[ v^2 = 50^2 - 2 \cdot 10 \cdot 55 \]
\[ v^2 = 2500 - 1100 \]
\[ v^2 = 1400 \]
\[ v = \sqrt{1400} \]
\[ v = \sqrt{100 \cdot 14} = 10\sqrt{14} \approx 37.42 \, \text{m/s} \]

### Step 4: Conclusion
The velocity of the body when it is at a height of 70 m above the ground is approximately \( 37.42 \, \text{m/s} \).

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