A body is thrown horizontally with a velocity of v m/s from the top of a tower of height 2h reaches the ground in 't' seconds. If another body double the mass is thrown horizontally with a velocity 5v m/s from the top of another tower of height 8h it reaches the ground in a time of

Options

A. 2t
B. 4t
C. 6t
D. 8t

Answer

To solve the problem step by step, we will analyze the motion of the two bodies thrown from the towers.

### Step 1: Understanding the time of flight for the first body
The first body is thrown horizontally from a height of \(2h\) with a velocity of \(v\) m/s. The time of flight \(t\) for a body thrown horizontally from a height \(h\) is given by the formula:

\[ t = \sqrt{\frac{2h}{g}} \]

For the first body, the height is \(2h\):

\[ t = \sqrt{\frac{2(2h)}{g}} = \sqrt{\frac{4h}{g}} = 2\sqrt{\frac{h}{g}} \]

### Step 2: Relating the time of flight to the given time
From the above calculation, we have:

\[ t = 2\sqrt{\frac{h}{g}} \]

### Step 3: Finding the time of flight for the second body
The second body is thrown horizontally from a height of \(8h\) with a velocity of \(5v\) m/s. We will denote the time of flight for the second body as \(t'\). Using the same formula for the time of flight:

\[ t' = \sqrt{\frac{2(8h)}{g}} = \sqrt{\frac{16h}{g}} = 4\sqrt{\frac{h}{g}} \]

### Step 4: Expressing \(t'\) in terms of \(t\)
Now we can express \(t'\) in terms of \(t\):

Since we have already established that:

\[ t = 2\sqrt{\frac{h}{g}} \]

We can substitute this into the equation for \(t'\):

\[ t' = 4\sqrt{\frac{h}{g}} = 2(2\sqrt{\frac{h}{g}}) = 2t \]

### Conclusion
Thus, the time taken by the second body to reach the ground is:

\[ t' = 2t \]

### Final Answer
The second body reaches the ground in \(2t\) seconds.

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