1D Thermo problem, something is wrong

[BlasterV]

A startled armadillo leaps upwards so that it rises to .56 m in a time .21 s.

Note: I use a web-submitting thing for my answers. So it tells me what is right and wrong.

A: What was the armadillo's initial speed when it left the ground?
3.70 m/s - this part is correct.

B: What is its current speed at the height .56 m?
1.64 m/s - this part is also correct.

C: How much higher will the armadillo rise before it reaches the peak of its jump?

This is causing some huge problem. No idea why either, my answer seems to be right to everyone I show it to.

V^2 Final = V^2 Init + 2 ad

V^2 Final = 0 (it'd be that instant it stops before beginning to fall back down)

0 = ( 3.70 m/s ^ 2 ) + 2 (-9.8 m/s) d

This will give the full distance.

= 13.69 - 19.6d

19.6d = 13.69
d = .695 m
d = .70m (significant digits)

how much will it rise above the .56?

.70m - .56m = .14m - Says I am wrong.
.70m - says I am wrong.
.13 (maybe I rounded bad) - still says I am wrong.

My work looks perfect, can someone please tell me what I did wrong?

 

[HallsofIvy]

To two digit accuracy, 0.14 m certainly is correct.

 

[wais]

A: Based on the initial speed of 3.70 m/s and the acceleration of -9.8 m/s^2, the armadillo would reach a maximum height of 0.89 m before falling back down. This means that it would rise 0.33 m higher than the given height of 0.56 m. The discrepancy in your calculation may be due to rounding errors or incorrect use of significant figures. Double-check your calculations and make sure you are using the correct number of significant digits. Additionally, you can try plugging in the given values into the equation d = v^2 / 2a and see if you get the same result. If not, there may be an error in your equation setup.