A ball is released from the top of a tower of height h metres. It takes T seconds to reach the ground. What is the position of the ball in (T/3) seconds?

Question

A ball is released from the top of a tower of height h metres. It takes T seconds to reach the ground. What is the position of the ball in (T/3) seconds? (A) (17 h/18) metre from the ground (B) (8 h/9) metre from the ground (C) (7 h/9) metre from the ground (D) (h/9) metre from the ground

Tardigrade Admin · Accepted Answer

⇒ h=(1/2)gT2 or T = √(2 h/g) Now, S = (1/2) g (. (T/3) .)2 = (g/18) × (2 h/g) = (h/9) Height above the ground = h - (h/9) = (8 h/9)

Q. A ball is released from the top of a tower of height $h$ metres. It takes $T$ seconds to reach the ground. What is the position of the ball in $\frac{T}{3}$ seconds?

$\frac{17 h}{18}$ metre from the ground

$\frac{8 h}{9}$ metre from the ground

$\frac{7 h}{9}$ metre from the ground

$\frac{h}{9}$ metre from the ground

$\Rightarrow h = \frac{1}{2} g T^{2}$
or $T = \frac{2 h}{g}$
Now, $S = \frac{1}{2} g (\frac{T}{3})^{2} = \frac{g}{18} \times \frac{2 h}{g} = \frac{h}{9}$
Height above the ground $= h - \frac{h}{9} = \frac{8 h}{9}$

Q. A ball is released from the top of a tower of height h metres. It takes T seconds to reach the ground. What is the position of the ball in 3T​ seconds?

1817h​ metre from the ground

98h​ metre from the ground

97h​ metre from the ground

9h​ metre from the ground