Question

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)

Answer 1

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 2

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