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A 0.11 kg ball of dough is thrown straight up into the air with an initial...

A 0.11 kg ball of dough is thrown straight up into the air with an initial speed of 17 m/s. The acceleration of gravity is 9.8 m/s 2 . What is its momentum halfway to its maximum height? Answer in units of kg m/s.

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Answer #1

here,

the initial speed , u = 17 m/s

the maximum height gained , hm = u^2 /( 2g)

hm = 17^2 /(2*9.81) m

hm = 14.73 m

the height at halfway to its maximum height , h = hm/2

h = 7.36 m

let the speed at this height be v

using third equation of motion

v^2 - u^2 = - 2 * g * h

v^2 - 17^2 = - 2 * 9.81 * 7.36

solving for v

v = 12.02 m/s

the momentum at this height , P = m * v

P = 0.11 * 12.02 kg.m/s

P = 1.32 kg.m/s

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