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 in T/3 seconds?
A. h/9 meter from the ground.
B. 7h/9 meter from the ground.
C. 8h/9 meter from the ground.
D. 17h/18 meter from the ground.

Answer 1

Hint : The acceleration acting on the ball will be g as it is just released from the top of the tower. Now apply the equation of motion to the values given. Now subtract the distance traveled from the total distance to find the distance from the ground.

Complete step by step solution :
The acceleration of the ball is g ,
Its initial velocity U=0,
We know that distance s is ,
$s=ut+\dfrac{1}{2}a{{t}^{2}}$
But u=0, and acceleration is equal to g so,
$\ \,\text{or}\ \,h=\dfrac{1}{2}g{{T}^{2}}$ $(U=0)$
Therefore, in T/3 seconds,
$h’ =\dfrac{1}{2}g{{(\dfrac{T}{3})}^{2}}$
On solving the equation we get,
$\Rightarrow h’ =\dfrac{1}{2}\times \dfrac{g{{T}^{2}}}{9}=\dfrac{h}{9}$
Now subtracting the distance travelled in T/3 seconds from the total distance of the tower,
$=h-\dfrac{h}{9}=\dfrac{8h}{9}$

So, the correct answer is “Option C”.

Additional Information:
The acceleration due to gravity is a force that pulls objects towards itself. Its value is 9.8m/s$^{2}$ . It is denoted by the letter ‘g’.
Any equation that describes the velocity of an object with respect to time is known as the equation of motion.
Initial velocity is the velocity from which a body starts to get accelerated.

Note : Always take acceleration due to gravity value as 9.8m/$s^2$ unless it's told. The value we get after putting the time as T/3 is the distance traveled. To get the position of the ball we have to subtract it from the total height.