A body rises vertically upward to a height of 100 m, in 5 seconds, then comes back to the same position after another 5 s. Find the distance travelled in m.

To find the distance traveled by the body, we need to consider the two parts of its motion separately: the upward motion and the downward motion.

1. Upward motion: The body rises vertically upward to a height of 100 m in 5 seconds. We can use the equation of motion for uniformly accelerated motion:

s = ut + (1/2)at^2

where s is the distance traveled, u is the initial velocity, t is the time taken, and a is the acceleration.

In this case, the initial velocity (u) is 0 m/s because the body starts from rest. The time taken (t) is 5 seconds, and the acceleration (a) can be calculated using the equation:

a = (v - u) / t

where v is the final velocity. Since the body comes back to the same position, the final velocity is also 0 m/s.

Therefore, the acceleration (a) is:

a = (0 - 0) / 5 = 0 m/s^2

Now, we can calculate the distance traveled (s) during the upward motion:

s = 0 * 5 + (1/2) * 0 * (5^2) = 0 + 0 = 0 m

2. Downward motion: The body comes back to the same position after another 5 seconds. Since the body is in free fall, the acceleration is equal to the acceleration due to gravity (g), which is approximately 9.8 m/s^2.

Using the equation of motion, we can calculate the distance traveled during the downward motion:

s = ut + (1/2)at^2

Here, the initial velocity (u) is 0 m/s, the time taken (t) is 5 seconds, and the acceleration (a) is -9.8 m/s^2 (negative because it is in the opposite direction to the upward motion).

s = 0 * 5 + (1/2) * (-9.8) * (5^2) = 0 + (-24.5) = -24.5 m

Since distance cannot be negative, we take the absolute value of -24.5 m:

| -24.5 | = 24.5 m

Therefore, the distance traveled during the downward motion is 24.5 m.

To find the total distance traveled, we add the distances traveled during the upward and downward motions:

Total distance = 0 m + 24.5 m = 24.5 m

Hence, the body traveled a total distance of 24.5 meters.