Calculating Acceleration of Locust Jumping 4 cm

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The discussion centers on calculating the acceleration of a locust that jumps 4 cm with an initial speed of 340 cm/s. The user initially calculated an acceleration of 145 m/s² using the formula (V_f)² = (V_i)² + 2as, with V_f as 3.4 m/s and V_i as 0. However, the solutions manual indicates the correct answer is 100 m/s². The user seeks clarification on the discrepancy between their calculation and the manual's answer.
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A locust, extending its hind legs over a distance of 4 cm, leaves the ground at a speed of 340 cm/s. Determine its acceleration, presuming it to be constant.

When I solved that, got an acceleration of 145 m/s^2 but the answer in the solutions manual is 100 m/s^2.
I used this formula :
(V_f)^2=(V_i)^2+2as
While, s=.04, V_f=3.4 m/s and V_i=0.
What's wrong ?
 
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Omid said:
What's wrong?

The solutions manual.
 
Really, it is a good news.
Thanks
 

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