Question

A body is released from the top of a towerof height 'h' metre . It takes ' T ' seconds to reach ground . Where is it at ' (T/2) ' seconds ?. A). (h/4)m. B). (h/2)m. C). (3h/4)m. D). depend upon its mass​

Answer 1

Answer:-

H = h/4 metre

Option → A

Given :-

Height of tower = h metre

Time taken to reach the ground = T seconds

To find :-

The position after T/2 seconds.

Solution :-

When a body is released from height it's initial velocity (u = 0)

and a = + g.

Now,

Using equation of motion under the gravity,

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

let after t/2 the body will be at H metre.

→$H = 0 \times T + \dfrac{1}{2}gT^2$

take T = T /2

→$H = \dfrac{1}{2}\times g (\dfrac{T}{2}))^2$

→$H = \dfrac {1}{2}\times g\dfrac{T^2}{4}$

→$H =\dfrac{1}{4} \times\dfrac{1}{2}gT^2$

→$H= \dfrac{h}{4}m$

hence, after time t/2 body will be at h/4 m.

Answer 2

Answer:

Option A is correct answer.

H = h/4 metre.

Explanation:

Given Problem:

A body is released from the top of a towerof height 'h' metre . It takes ' T ' seconds to reach ground . Where is it at ' (T/2) ' seconds ?. A). (h/4)m. B). (h/2)m. C). (3h/4)m. D). depend upon its mass​.

Solution:

To Find:

The position after T/2 seconds.  

--------------------

Method:

Given that,

Height of tower = h metre.

Time taken to reach the ground = T seconds.

We know that,

When a body is released from height it's initial velocity (u = 0)  and a = + g.

Now,

We know that,

★Equation of motion under the gravity,★

★$h = ut+\dfrac{1}{2}gt^2$★

Now,

Let after t/2 the body will be at H metre.

$\implies\ H = 0 \times T+\dfrac{1}{2} gt^2$

Now,

Take T = T/2

$\implies\ H=\dfrac{1}{2} \times g(\dfrac{T}{2})^2$

$\implies\ H = \dfrac{1}{2} \times g\dfrac{T^2}{4}$

$\implies\ H = \dfrac{1}{4} \times \dfrac{1}{2}gT^2$

$\implies\ H = \dfrac{h}{4}meter$

Hence,

It implies that the after time t/2 body will be at h/4 m.