A body is projected with a velocity `50 "ms"^(-1)`. Distance travelled in 6th second is `[g=10 ms^(-2)]`

Options

A. 5m
B. 10m
C. 15m
D. 20m

Answer

To solve the problem of finding the distance traveled by a body projected upwards with an initial velocity of \(50 \, \text{m/s}\) during the sixth second, we can follow these steps:

### Step 1: Understand the motion
The body is projected upwards with an initial velocity \(u = 50 \, \text{m/s}\). The acceleration due to gravity \(g = 10 \, \text{m/s}^2\) acts downwards, so we will take it as negative when calculating the distance traveled upwards.

### Step 2: Find the time to reach maximum height
To find the time taken to reach the maximum height, we use the first equation of motion:
\[ v = u + at \]
At maximum height, the final velocity \(v = 0\). Thus,
\[ 0 = 50 - 10t \implies 10t = 50 \implies t = 5 \, \text{s} \]
So, the body reaches its maximum height at \(t = 5 \, \text{s}\).

### Step 3: Calculate the distance traveled in the 5th and 6th seconds
The distance traveled in the \(n\)-th second can be calculated using the formula:
\[ s_n = u + \frac{1}{2} a (2n - 1) \]
For the 6th second (\(n = 6\)):
\[ s_6 = u + \frac{1}{2} a (2 \cdot 6 - 1) = 50 + \frac{1}{2} (-10)(11) = 50 - 55 = -5 \, \text{m} \]
This negative value indicates that the body is falling back down during the 6th second.

### Step 4: Calculate the distance traveled in the 5th second
For the 5th second (\(n = 5\)):
\[ s_5 = u + \frac{1}{2} a (2 \cdot 5 - 1) = 50 + \frac{1}{2} (-10)(9) = 50 - 45 = 5 \, \text{m} \]

### Step 5: Calculate the distance traveled in the 6th second
The distance traveled in the 6th second is given by:
\[ \text{Distance in 6th second} = s_6 - s_5 \]
Since the body is falling back down, we can also directly calculate it as:
\[ \text{Distance in 6th second} = s_6 = -5 \, \text{m} \]

### Conclusion
The distance traveled in the 6th second is \(5 \, \text{m}\) downwards.