Physics, asked by Masuka549, 1 year ago

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 gadakhsanket
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...

Answered by karthikmamididr
80

Answer:

 \sqrt{8 {m}^{2}gh }

Explanation:

change in momentum = force × time

impulse = change in momentum

I = F× T------- equation 1

force = mass × gravity

time \: of \: flight =  \frac{2u}{g}

substitute the values in equation 1

I = F × T

i = mg \times  \frac{2u}{g}

i = 2mu -  -  -  -  - equation \: 2

 {v}^{2}  -  {u}^{2}  = 2as

v = 0

a = -g

s = h

 u =  \sqrt{2gh}

substitute the value of u in equation 2

i = 2mu

i = 2m \sqrt{2gh}

i =  \sqrt{ {(2m)}^{2}2gh }

i =  \sqrt{4 {m}^{2}2gh }

i =  \sqrt{8 {m}^{2}gh }

I hope the answer will help you

thank you

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