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An umpire tosses up a coin with a speed of 10 m/s. If his hand is 1 m above the ground,...
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Let us say the datum is at the level of umpires hand.
Using equations of motion under gravity for the travel of coin comes to max height.
`uarrv^2 = u^2+2gS`
`S = (v^2-u^2)/(2g)`
`S = (10^2-0)/(2xx9.81)`
`S = 5`
So it will come another 5m to the datum level and go further 1m down to reach the ground.
Total distance travelled by coin = 5+5+1 = 11m.
So the answer is d.
Posted by jeew-m on May 2, 2013 at 11:51 AM (Answer #1)
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