Calculate Height of Ball After Leaving Hand: 7.2 m

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The problem involves calculating the height a 200-g ball reaches after being thrown with an upward force of 9.4 N for 0.32 seconds. The net acceleration is calculated as 47 m/s², leading to an initial velocity of 15 m/s. Using the kinematic equation for height, the calculated height is 11.5 m, which differs from the book's answer of 7.2 m. The discussion raises the question of whether to use the upward acceleration or the net acceleration for the calculation. Clarifying this distinction is crucial for arriving at the correct answer.
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Homework Statement


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.


Homework Equations


F = ma
v = v0 + at
y - y0 = v^2/2g

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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