Question

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 of the ball in T/3 second -

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Answer 1

Answer:

Distance of the ball from the top of the tower = h/9

Distance of the ball from the ground = 8h/9

Step-by-step explanation:

We've been given that, a ball is released from the top of the tower of height 'h' meters. It takes 'T' seconds to hit the ground. [Kindly check the diagram]

With this information, we've been asked to find out the position of the ball (distance) covered by it in T/3 seconds.

According to the second law/equation of motion,

$\sf \Longrightarrow s = ut + \dfrac{1}{2}gt^2$

At t = 'T' seconds,

• s = Distance covered = h

• u = Initial velocity = 0

• t = Time taken = T

• g = Acceleration due to gravity = g

$\sf \Longrightarrow h = 0(T) + \dfrac{1}{2}g(T)^2$

$\sf \Longrightarrow \textsf{\textbf{h}} =\dfrac{\textsf{\textbf{1}}}{\textsf{\textbf{2}}}\textsf{\textbf{gT}}^2 \ \ \longrightarrow Relation(1)$

At t = 'T/3' seconds we have,

• s = Distance covered = h'

• u = Initial velocity = 0

• t = Time taken = T/3

• g = Acceleration due to gravity = g

$\sf \Longrightarrow s = ut + \dfrac{1}{2}gt^2$

$\sf \Longrightarrow h' = (0)(t/3) + \dfrac{1}{2}g\bigg(\dfrac{T}{3}\bigg)^2$

$\sf \Longrightarrow h' = \dfrac{1}{2}g \times \dfrac{T^2}{9}$

$\sf \Longrightarrow h' = \dfrac{1}{2}g \times T^2 \times \dfrac{1}{9}$

$\sf \Longrightarrow h' = \dfrac{1}{2}gT^2 \times \dfrac{1}{9}$

From Relation(1), we know that (1/2)gT² = h, on substituting it above we get,

$\sf \Longrightarrow h' = h \times \dfrac{1}{9}$

$\sf \Longrightarrow \textsf{\textbf{h'}} = \dfrac{\textsf{\textbf{h}}}{\textsf{\textbf{9}}} \ \ \longrightarrow Relation(2)$

∴ The distance of the ball at T/3 from the tower is h/9.

To calculate the position of the ball from the ground, we subtract the height of the tower (h) from the distance covered by the ball in t/3 seconds (h/9).

$\sf \Longrightarrow Distance \ from \ the \ ground = h - \dfrac{h}{9}$

$\sf \Longrightarrow Distance \ from \ the \ ground = \dfrac{9h - h}{9}$

$\sf \Longrightarrow \textsf{\textbf{Distance \ from \ the \ ground}} = \dfrac{\textsf{\textbf{8h}}}{\textsf{\textbf{9}}}$

∴ The distance of the ball at T/3 from the ground is 8h/9.

Answer 2

Given :-

A ball is released from the top of a tower of height h meters. It takes T second to reach the ground.

To Find :-

Position of the ball in T/3 second -

Solution :-

As the ball starts from rest. Then, initial velocity is 0 m/s.

We know that

$\sf h = ut + \dfrac{1}{2}gt^2$

Here,

h = height

u = initial velocity

t = time

g = acceleration due to gravity

◼ I n C a s e 1 : -

Time = T second

u = 0 m/s (As it starts from rest)

g = 10 m/s²

$\sf h_{1} = 0\times T + \dfrac{1}{2}\times 10\times T^2$

$\sf h_{1} = 0 + 5\times T^2$

$\sf h_{1} = 5T^2$

◼ I n C a s e 2 : -

Time = T/3 second

u = 0 m/s

g = 10 m/s²

$\sf h_{2} = 0\times\dfrac{T}{3}+\dfrac{1}{2}\times10\times\bigg(\dfrac{T}{3}\bigg)^2$

$\sf h_{2} = 0 + 5 \times\dfrac{T^2}{9}$

$\sf h_{2}=5\times\dfrac{T^2}{9}$

$\sf h_{2}=\dfrac{5T^2}{9}$

$\sf h_{2}=\dfrac{h_1}{9}$

Now, Finding Position of ball in T/3 second

$\sf Position_{\frac{T}{3}\;sec} = h_1 - \dfrac{h_1}{9}$

$\sf Position_{\frac{T}{3}\;sec}=\dfrac{9h_1-h_1}{9}$

$\sf Position_{\frac{T}{3}\;sec} =\dfrac{8h}{9}\;m$