Question

8. You drop a 2.9 kg foam ball from a very high building, Your data indicates that it reaches a terminal vélocity s a s after you drop it. Its terminal 8a) What is the terminal velocity of the ball? velocity is equal to ½ of the velocity it would have after 50s if there was no air resistance

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Answer #1

using 1st equation of motion to find velocity before it hit the ground in absence of air

v = u + at

since the object dropped, so, u = 0

v = 9.8* 5 = 49 m/s

now given that

terminal velocity = 0.5 * v = 8.5*49 = 24.5 m/s

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Do comment in case any doubt, will reply for sure . Goodluck

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