Two bodies,one held 0.98m vertically above other,are released simultaneously and fall freely under gravity.After 2 seconds,their relative separation is?
since the acc on them is constant (9.8 m/s) hence till the time the body vertically high covers o.98 m the second body wuld have been displaced 0.98mts downward n iff the downfall continues for infinity sio the distance between the two bodies wuld always be 0.98 m fo the nth second
Education
Let the body 0.98 m above be B and the lower body be B*.
THEREFORE:- for B
u= 0m/s
g=(-9.8)m/s² ( - sign because of sign convention)
t=2 s
So By - S= ut+½at²
S=½(-9.8)(4)
S= -19.6 m.-------1)
For second body B*
u=0
a=-9.8m/s²
t=2 s
S*=ut+½at²
S*=½(-9.8)(4)
S*= (-19.6m)
Therefore S*=S. SO initial separation is equal to final separation = 0.98m. This is the answer.
To determine the relative separation between the two bodies after 2 seconds, we can use the equations of motion for a freely falling object under gravity. The key equation we need is:
h = ut + (1/2)at^2
where:
h is the height or separation of the bodies,
u is the initial velocity (which is 0 since both bodies are released),
a is the acceleration due to gravity (approximately 9.8 m/s^2),
t is the time (2 seconds in this case).
First, let's find the separation for each body after 2 seconds.
For the body held 0.98m vertically above the other:
h1 = (0 * 2) + (0.5 * 9.8 * 2^2)
For the body below:
h2 = (0 * 2) + (0.5 * 9.8 * 2^2)
Now, let's find the relative separation between the two bodies by subtracting these values:
Relative separation = h2 - h1
Substituting the values, we get:
Relative separation = [(0 * 2) + (0.5 * 9.8 * 2^2)] - [(0 * 2) + (0.5 * 9.8 * 2^2)]
Simplifying:
Relative separation = 0
Therefore, the relative separation between the two bodies after 2 seconds is 0 meters.