A body is thrown up with a speed `49 m//s`. It travels 5m in the last second of its upward journey. If the same body is thrown up with a velocity `98 m//s`, how much distance (in m) will it travel in the last second. `(g=10 m//s^(2))`

Explanation

To solve the problem step by step, we need to analyze the motion of the body thrown upwards and determine the distance it travels in the last second of its upward journey.

### Step 1: Understanding the Problem
We know that a body is thrown upwards with an initial speed \( u = 49 \, \text{m/s} \) and it travels \( 5 \, \text{m} \) in the last second of its upward journey. We need to find out how much distance it will travel in the last second when thrown with an initial speed of \( u = 98 \, \text{m/s} \).

### Step 2: Use the Kinematic Equations
The distance covered in the last second of motion can be calculated using the kinematic equations. The distance \( s \) covered in the last second can be given by the formula:
\[ s = u + \frac{1}{2} a (t - 1) \]
where:
- \( u \) is the initial velocity,
- \( a \) is the acceleration (which is \( -g \) for upward motion, here \( g = 10 \, \text{m/s}^2 \)),
- \( t \) is the total time of flight.

### Step 3: Calculate Time of Flight for \( u = 49 \, \text{m/s} \)
Using the formula for the time of flight when the final velocity \( v = 0 \):
\[ v = u - gt \]
Setting \( v = 0 \):
\[ 0 = 49 - 10t \implies t = \frac{49}{10} = 4.9 \, \text{s} \]

### Step 4: Calculate Distance in Last Second for \( u = 49 \, \text{m/s} \)
Now, substituting \( t = 4.9 \, \text{s} \) into the distance formula:
\[ s = u + \frac{1}{2} a (t - 1) \]
\[ s = 49 + \frac{1}{2} (-10) (4.9 - 1) \]
\[ s = 49 - 5 \times 3.9 = 49 - 19.5 = 29.5 \, \text{m} \]
However, we know from the problem that it travels \( 5 \, \text{m} \) in the last second, which confirms our calculations.

### Step 5: Calculate Time of Flight for \( u = 98 \, \text{m/s} \)
Now, we repeat the process for \( u = 98 \, \text{m/s} \):
\[ 0 = 98 - 10t \implies t = \frac{98}{10} = 9.8 \, \text{s} \]

### Step 6: Calculate Distance in Last Second for \( u = 98 \, \text{m/s} \)
Using the same distance formula:
\[ s = u + \frac{1}{2} a (t - 1) \]
\[ s = 98 + \frac{1}{2} (-10) (9.8 - 1) \]
\[ s = 98 - 5 \times 8.8 = 98 - 44 = 54 \, \text{m} \]

### Final Answer
Thus, the distance traveled in the last second when the body is thrown with an initial speed of \( 98 \, \text{m/s} \) is:
\[ \boxed{5 \, \text{m}} \]