A man of mass 2m with a bag of mass m is standing on a tower of height H .There is a pond at x distance from the tower. He jumps from the tower with the bag. After free falling for a distance/height h he throws the bag towards the tower with a velocity z to fall into the pond. Find z.
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after falling from height H , man stop at distance h . Let speed of man is u and speed of bag is z (v) in horizontal direction .
due to external for is zero .
momemtu must be conserve at h distance from the ground .
e.g mass of man × velocity of man = mass of bag × velocity of bag
=> 2mu = mz
=> 2u = z ___________(1)
see the figure for understanding , here in figure v is z.
h = ut +1/2 at²
h = 0× t + 1/2gt² = 1/2gt²
t = √(2h/g)
and horizontal distance covered by man
x = ut
x = u√(2h/g)
u = x√(g/2h) put this in eqn (1)
z = 2x(g/2h)½
hence, z = 2x{g/2h)½
due to external for is zero .
momemtu must be conserve at h distance from the ground .
e.g mass of man × velocity of man = mass of bag × velocity of bag
=> 2mu = mz
=> 2u = z ___________(1)
see the figure for understanding , here in figure v is z.
h = ut +1/2 at²
h = 0× t + 1/2gt² = 1/2gt²
t = √(2h/g)
and horizontal distance covered by man
x = ut
x = u√(2h/g)
u = x√(g/2h) put this in eqn (1)
z = 2x(g/2h)½
hence, z = 2x{g/2h)½
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abhi178:
I hope it'll correct answer .
um....actually the distance h is not from the ground but from the top of the tower
okay then answeer will be z = 2x (g/2 (H-h)}^1/2
Thank you so much!!
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