A body is projected with a velocity `60 ms^(-1)` vertically upwards the distance travelled in last second of its motion is `[g=10 ms^(-1)]`

Options

A. 35 m
B. 45 m
C. 55 m
D. 65 m

Answer

To solve the problem of finding the distance traveled in the last second of a body projected vertically upwards with an initial velocity of \(60 \, \text{m/s}\), we will follow these steps:

### Step 1: Calculate the Time of Flight
The total time of flight \(T\) for a body projected upwards can be calculated using the formula:
\[ T = \frac{2u}{g} \]
where \(u\) is the initial velocity and \(g\) is the acceleration due to gravity.

Given:
- \(u = 60 \, \text{m/s}\)
- \(g = 10 \, \text{m/s}^2\)

Substituting the values:
\[ T = \frac{2 \times 60}{10} = \frac{120}{10} = 12 \, \text{s} \]

### Step 2: Calculate the Distance Traveled in the Last Second
To find the distance traveled in the last second (which is the 12th second), we can use the formula for the distance traveled in the \(n^{th}\) second:
\[ d_n = u + \frac{a}{2} \cdot (2n - 1) \]
where:
- \(u\) is the initial velocity,
- \(a\) is the acceleration (which is \(-g\) for upward motion),
- \(n\) is the second we are interested in.

For the 12th second:
- \(n = 12\)
- \(a = -10 \, \text{m/s}^2\)

Substituting the values into the formula:
\[ d_{12} = 60 + \frac{-10}{2} \cdot (2 \times 12 - 1) \]
\[ d_{12} = 60 - 5 \cdot (24 - 1) \]
\[ d_{12} = 60 - 5 \cdot 23 \]
\[ d_{12} = 60 - 115 = -55 \, \text{m} \]

### Conclusion
The distance traveled in the last second of the motion is \(-55 \, \text{m}\). The negative sign indicates that the body is moving downwards at this point in its motion.