A student throws a water balloon vertically downward from the top of a building. The balloon leaves the thrower's hand with a speed of 6.35m/s . Air resistance may be ignored, so the water balloon is in free fall after it leaves the thrower's hand.(adsbygoogle = window.adsbygoogle || []).push({});

A.) What is its speed after falling for a time 1.65s ? Take the free fall acceleration to be g=9.80m/s^2 .

my work:

using formula: v(t) = v(0) + at

v(t) = (6.35) + (-9.8)(1.65) = -9.82

B.) How far does it fall in a time of 1.65s ? Take the free fall acceleration to be 9.80 m/s^2.

my work: x(t) = x(0) + v(0)t + 1/2at^2

x(0) = (6.35)(1.65) + 1/2(-9.8)(1.65)^2

x = 2.86

C.) What is the magnitude of its velocity after falling a distance 10.9m ? Take the free fall acceleration to be g=9.80 m/s^2.

my work:

v^2 = v(0)^2 + 2a(x-x(0))

v^2 = +2(-9.8)(10.9)

v= 14.616

ok, i got them all wrong. what am i doing wrong? i tried everything that i know and cant get it right, please help

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# Homework Help: Free fall question

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