Question

derive the expression for the time taken by a body dropped from a height h to reach at earth​

Answer 1

Given:

A body is dropped from height h .

To find:

Time taken to reach earth surface?

Calculation:

• Let the time taken be t.
• Also, since the body is dropped, the initial velocity will be zero.

Applying EQUATIONS OF KINEMATICS:

$h = ut + \dfrac{1}{2} a {t}^{2}$

$\implies \: h = 0(t) + \dfrac{1}{2} g {t}^{2}$

$\implies \: h = \dfrac{1}{2} g {t}^{2}$

$\implies \: {t}^{2} = \dfrac{2h}{g}$

$\implies \: t = \sqrt{\dfrac{2h}{g} }$

So, time taken is √(2h/g) .

Answer 2

Answer:

t² = √2h/g

Explanation:

Let, time taken be t.

initial velocity will be 0.

h=ut + 1/2 at²

h=0(t) + 1/2gt²

h=1/2gt²

t²=2h/g

t²=√2h/g

hope it helps you!!