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Employing the well-known physics formula h=gt^2/2, we are required to display the theoretical time it would take for an object to reach the ground. I isolated variable t=sqrt(2h/g) and inserted formula =SQRT(QUOTIENT(2*B3,$B$1)), yet the values reaped were incorrect.

Source data: enter image description here

  • Expected valid results for t = 1 to t = 8 are respectively .45, .64, .78, .90, 1.01, 1.11, 1.19 and 1.27.
  • What do you mean by "values reaped were incorrect"? Your formula isn't even valid, so you would only get error message and not incorrect values. Please provide a valid formula, the source data, and the expected valid results. – Ron Rosenfeld Jun 7 at 1:00
  • I used the formula of =SQRT(QUOTIENT(2*B3,$B$1)), exactly what you suggested in your answer, but output values obtained still are incorrect. No error message was received in any of the cells. – TheQuantumObsession Jun 7 at 1:13
  • Please provide a valid formula, the source data, and the expected valid results. Do this by editing your answer. – Ron Rosenfeld Jun 7 at 1:15
  • See my revised answer. What was originally my 2nd formula returns your desired results. – Ron Rosenfeld Jun 7 at 1:40
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Try this instead:

=SQRT(2*B3/$B$1)

enter image description here

The above results are calculated to Excel's precision limit and then rounded to two decimals by the cell formatting option. There is no easy way to exactly mimic your stated desired results.


The QUOTIENT function returns only the Integer part of the division, which is why using it in your formula will return 0's or 1's, depending on the value of t

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  • In all practicality, I would only need to round to 2 decimal places, so I'm perfectly fine with such. I wasn't aware of the quotient function only returning the integer part of the division, so thank you for your following up. – TheQuantumObsession Jun 7 at 1:59
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You may try these:

=SQRT(TRUNC(2*B3/$B$1))
=SQRT(ROUND((2*B3/$B$1),0))
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