A ball is released from the top of a tower of height h meters. It takes T second to reach the ground. What is the position f the ball in T/3 second

Question

mathslover · Answer

I did like this : 
$$\large{T =\sqrt{\frac{2h}{g}}}$$

$$\large{v^2=u^2+2gh'}$$
$$\large{g^2(T^2)=2gh'}$$
$$\large{g^2*\frac{2h}{g}*\frac{1}{9}=2gh'}$$
$$\large{\frac{h}{9}=h'}$$

anonymous · Answer

we can directly do it like this also rit....h=1/2*g*T/3*T/3===>h/9

mathslover · Answer

but the answer is 8h / 9

ajprincess · Answer

When u measure from the top h'=h/9 . it is h-h/9=8h/9 from the ground.

ash2326 · Answer

@mathslover  In your coordinate system
you took the top of tower as 0 and the ground as h
so when t=T/3 the ball is at h/9 from tower so, it'd be 8h/9 from the ground

mathslover · Answer

hmn thnks to both @ajprincess and @ash2326 .. i got the mistake now

anonymous · Answer

tnx guys^ ^

mathslover · Answer

:D this is good , that you also learnt something from this

thanks to all

ajprincess · Answer

U r welcome.:D

mathslover · Answer

:)