a body is released from the great height and falls freely towards the earth . another body is released from the same height , one second later . then the separation between the two bodies , two seconds after release of the second body is
free fall...starting from rest
d = 1/2 g t^2
... plug in the times, and find the difference in the distances
the 1st body will always be going faster than the 2nd, so the distance between them is always increasing
25.4
To find the separation between the two bodies two seconds after the release of the second body, we need to consider the motion of each body independently.
When a body is freely falling towards the Earth, it is subject to the force of gravity, which causes it to accelerate downwards at a constant rate. This acceleration due to gravity is approximately 9.8 m/s^2.
Let's assume that the initial height of both bodies is "h" (from the given information). The distance traveled by a freely falling body in time "t" can be calculated using the equation:
s = ut + (1/2)at^2
Where:
- s is the distance traveled
- u is the initial velocity (which is 0 since both bodies are initially at rest)
- a is the acceleration due to gravity
- t is the time
For the first body, which is released immediately, the time of travel is 2 seconds. Plugging in the values, we have:
s1 = (0)(2) + (1/2)(9.8)(2^2)
s1 = 0 + (1/2)(9.8)(4)
s1 = 0 + 19.6
s1 = 19.6 meters
For the second body, which is released one second later, the time of travel is 1 second. Plugging in the values, we have:
s2 = (0)(1) + (1/2)(9.8)(1^2)
s2 = 0 + (1/2)(9.8)(1)
s2 = 0 + 4.9
s2 = 4.9 meters
Finally, to find the separation between the two bodies two seconds after the release of the second body, we subtract the distance traveled by the second body from the distance traveled by the first body:
Separation = s1 - s2
Separation = 19.6 - 4.9
Separation = 14.7 meters
Therefore, the separation between the two bodies two seconds after the release of the second body is 14.7 meters.