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Answer:
The height of the cliff is 90.60 meters.
Explanation:
It is given that,
Initial horizontal speed of the stone, u = 10 m/s
Initial vertical speed of the stone, u' = 0 (as there is no motion in vertical direction)
The time taken by the stone from the top of the cliff to the bottom to be 4.3 s, t = 4.3 s
Let h is the height of the cliff. Using the second equation of motion in vertical direction to find it. It is given by :
h = 90.60 meters
So, the height of the cliff is 90.60 meters. Hence, this is the required solution.
Answer:
m1/6 ( c )
Explanation:
since all the balls starts having the same momentum after the two collisions we will apply the principal of conservation of energy
After first collision
m1v = m1v1 + m2v2 --- ( 1 )
After second collision
m2v2 = m2v2 + m3v3 ---- ( 2 )
combining equations 1 and 2
m1v = m1v1 + m2v2 + m3v3 ----- ( 3 )
All balls moving at the same momentum ( p ) = m1v1 = m2v2 = m3v3
note ; 3p = m1v ∴ m3 = ----- ( 4 )
applying conservation of energy
3v = v1 + v2 + v3 ------- ( 5 )
also 3m1v1 = m1v = v1 = v/3 =
v2 + v3 = 8/3 v ----- ( 6 )
next eliminate V3 for equation 6 by applying conservation of energy and momentum
m1 = 2m2 ------ ( 7 )
now using p1 = p2 = m1v1 = 1/2 m1v1 hence v2 = 2v1 where v1 = 1/3 v
hence ; v2 = 2/3 v ------- ( 8 )
solving with equation 6 and 8
v3 = 2v ------ ( 9 ) ∴ v/v3 = 1/2 ---- ( 10 )
solving with equation 9 and 10
m3 = m1/3 * 1/2 = m1/6