Question 1090435
A body travels half of its path length in last second of its free fall. Calculate the height from which it falls ?
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Using h(t) = 16t^2:
d1 = 16t^2 --- time minus 1 second
d2 = 16(t+1)^2 --- total time falling
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d2 = 2d1
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16(t+1)^2 = 32t^2
t^2 + 2t + 1 = 2t^2
t^2 - 2t - 1 = 0
*[invoke solve_quadratic_equation 1,-2,-1]
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t = sqrt(2) + 1 seconds.
h = 16t^2
h = 16*(3 + 2sqrt(2)) = ~ 93.25 feet
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Note:  The t is exact.
The selection of units (feet, meters, etc) affects the result, as does the approximation of 16t^2.
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Also, the height cannot be determined unless it's assumed the object impacts after the calculated time.