A body is thrown up with a velocity 29.23m$s^{-1}$ distance travelled in lst second of upward motion is

During upward Motion a = -g

u = 29.23 m/s

Using Formula for distance in nth Sec 

D_n  = 4+ $\frac{a}{2}(2n-1)$

Time Ascent = $\frac{u}{g}$ = $\frac{29.23}{9.8} = 2.98 Sec$ 

S₃  = 29.23 - $\frac{9.8}{2}(2.98 \times 2 - 1)$

S₃ = 29.23 - 4.9 (5)

 S₃ = 29.23 -  24.304

S₃ = 4.926 m