A ball of mass m is thrown vertically upward from the ground and reaches a height h before momentarily coming to rest. If g is acceleration due to gravity, what is the impulse received by the body due to gravity force during its flight? (Neglect air resistance)
Answers
Answered by
32
Hii dear,
# Answer- I = -2m√(gh)
# Explaination-
When an object is thrown vertically upwards at height b and falls down,
displacement s = 2h
acceleration a = -g
By Newton 3rd kinematic eqn,
v^2 = u^2 + 2as
0 = u^2 + 2(-g)(2h)
u = √(4gh)
Impulse = Change in momentum
I = m(v-u)
I = m[0-√(4gh)]
I = -2m√(gh)
Hope that helped you...
# Answer- I = -2m√(gh)
# Explaination-
When an object is thrown vertically upwards at height b and falls down,
displacement s = 2h
acceleration a = -g
By Newton 3rd kinematic eqn,
v^2 = u^2 + 2as
0 = u^2 + 2(-g)(2h)
u = √(4gh)
Impulse = Change in momentum
I = m(v-u)
I = m[0-√(4gh)]
I = -2m√(gh)
Hope that helped you...
Answered by
80
Answer:
Explanation:
change in momentum = force × time
impulse = change in momentum
I = F× T------- equation 1
force = mass × gravity
substitute the values in equation 1
I = F × T
v = 0
a = -g
s = h
substitute the value of u in equation 2
I hope the answer will help you
thank you
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