Calculate Height of Ball After Leaving Hand: 7.2 m

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SUMMARY

The discussion centers on calculating the height a 200-gram ball reaches after being thrown with a constant upward force of 9.4 N for 0.32 seconds. The established answer is 7.2 meters, derived from the equations of motion. The net acceleration calculated is 47 m/s², leading to an initial velocity of 15 m/s. The final height calculation using the equation y - y0 = v0²/2g results in 11.5 meters, prompting a debate on whether to use the upward acceleration or the net acceleration in the calculations.

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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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