(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

In throwing a 200-g ball, one's hand exerts a constant upward force of 9.4 N for 0.32 s. How high does the ball rise after leaving the hand? Book answer is 7.2 m.

2. Relevant equations

F = ma

v = v0 + at

y - y0 = v^2/2g

3. The attempt at a solution

Fup + Fdn = Fnet

aup = Fup/m = 9.4 kg-m/s^2/200g*1000g/kg = 47 m/s^2

v0 = at = 47m/s^2 * 0.32s = 15 m/s

y-y0 = v0^2/2g = (15)^2/2 * 9.8 = 11.5 m

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# Homework Help: Force and motion

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