### Question Description

A ball of mass m is thrown straight up into the air with an initial speed v0. (a) Find an expression for the maximum height reached by the ball in terms of v0 and g. hmax =

(b) Using conservation of energy and the result of part (a), find the magnitude of the momentum of the ball at one-half its maximum height in terms of m and v0. p =

## Final Answer

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a)H maximum = Hmax

at maximum height V = 0 then gt = Vo ==> t = Vo/g

Hmax = (Vo)²/g - (Vo)²/(2g)

Hmax = (Vo)²/(2g)

b) conservation of energy means that energy, KE+PE, is a constant.

momentum=mass*velocity=mV

m(Vo)²/2 + 0 = mgH + mV²/2

V² = (Vo)² - 2gH

Hmax/2 = (Vo)²/(4g)

V² = (Vo)² - 2g(Vo)²/(4g)

V² = (Vo)² - (Vo)²/2

V² = (Vo)²/2 ==> V = ((Vo)√2)/ 2

mV = (m(Vo)√2)/2

NOTE: g is the gravity=9.8N

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