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A stone is dropped from height h and falls the last half of its distance in 4 secs....
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Starting its flight from height h, let the stone takes ` t_1` seconds to reach the halfway mark, and `t_2` seconds to reach the ground.
From the equations of motion, the final velocities at those two points are:
`v_1=sqrt(2gh/2)` , i.e. `sqrt(gh)`
Time to reach the velocities as mentioned are:
By the condition of the problem `t_2-t_1=4s`
Squaring both sides,
`rArr h=32.2*93.25483=3002.8` ft
and total time of flight, `t_2=sqrt(2h/g)`
Posted by llltkl on September 1, 2013 at 7:30 PM (Answer #1)
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