Question

un corp cade liber de la inaltimea h -176,4 m. ce distanta a parcurs el in ultima secunda inainte de a atinge pamintul ? dar in penultima??

Answer 1

a) $h=\frac{gt^2}{2}=\ \textgreater \ t=\sqrt{\frac{2h}{g}}=5,94 s$

$T=5,94-1=4,94s$
 
In ultima secunda corpul parcurge distanta 
$H=h-h'=h-\frac{gT^2}{2}=122,018=54,382m$

b) $T'=5,94-2=3,94 s$

In timpul T' corpul parcurge  distanta 

$D=\frac{gT'^2}{2}=77,618m$

Distanta pe care o parcurge in penultima secunda este 
 
$D'=h-H-D=44,4m$