Two bodies,one held 0.98m vertically above other,are released simultaneously and fall freely under gravity.After 2 seconds,their relative separation is?
4 answers
If they have landed, zero. If not, .98m.
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.
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.
1 answer