Question
A ball is released from the top of a tower of height $$h$$ meters. It takes $$T$$ seconds to reach the ground. What is the position of the ball at $$\frac{T}{3}$$ second-
A.
$$\frac{{8h}}{9}$$ meters from the ground
B.
$$\frac{{7h}}{9}$$ meters from the ground
C.
$$\frac{{h}}{9}$$ meters from the ground
D.
$$\frac{{17h}}{18}$$ meters from the ground
Answer :
$$\frac{{8h}}{9}$$ meters from the ground
Solution :
We know that
$$s = ut + \frac{1}{2}g{t^2},\,\,or\,h = \frac{1}{2}g{T^2}\left( {\because u = 0} \right)$$
now for $$\frac{T}{3}$$ second, vertical distance moved is given by
$$h' = \frac{1}{2}g{\left( {\frac{T}{3}} \right)^2} \Rightarrow h' = \frac{1}{2} \times \frac{{g{T^2}}}{9} = \frac{h}{9}$$
$$\therefore $$ position of ball from ground $$ = h - \frac{h}{9} = \frac{{8h}}{9}$$