Question

From a height 'h' a body reaches ground in 't' sec.Find the to cover h/2 in term gt

Answer 1

Answer:

Given:

An object takes 't' time to reach ground from a height 'h'.

To find:

The time to cover (h/2) in terms of 'g' , 't'

Concept:

We have to use the equations of Kinematics in order to solve this problem.

Calculation:

h = ut + ½gt²

=> h = (0 × t ) + ½gt²

=> h = ½gt² ...............(1)

1. Now to cover the upper (h/2) distance

s = ut + ½gT²

=> (h/2) = (0 × t) + ½gT²

=> h/2 = ½gT² ...............(2)

Comparing equation (1) and (2):

{h/(h/2)} = (½gt²)/(½gT²)

=> 2 =( t/T)²

=> T = (t/√2).

2. Now , time to cover lower (h/2) distance:

Let the time be denoted by T'.

So T' = t - T

=> T' = t - (t/√2)

=> T' = t ( 1- 1/√2)

=> T' = t {(√2-1)/√2}.

Answer 2

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By using the equations of Kinematics we have to find the answer →

h = u t + ½ g t²

→ h = (0 × t ) + ½gt²

→ h = ½gt² _______________(1)

1. Now to cover the upper (h/2) distance

s = u t + ½ g T²

→(h/2) = (0 × t) + ½gT²

→ h/2 = ½gT² _____________(2)

By Comparing equation (1) and (2) →

[h/(h/2)] = [(½gt²)/(½gT²)]

→2 =( t/T)²

→ T = (t/√2).

2. Now , time to cover lower (h/2) distance:

Let the time be denoted by ( T* )

Hence,

→T* = ( t - T )

→ T* = t - (t/√2)

→ T* = t ( 1- 1/√2)

→T* = t {(√2-1)/√2}