Question

A body falling from height 'h' takes t1 time to reach the ground. The time taken to cover the first half of the height is?

Answer 1

Use the second equation of motion.$h=\frac{1}{2}g{t_1}^{2}\\\frac{h}{2}=\frac{1}{2}gt^2\\\text{Put the value of h from first equation into second equation}\\\frac{1}{4}g{t_1}^{2}=\frac{1}{2}gt^2\\\frac{1}{2}{t_1}^{2}=t^2\\t=\frac{t_1}{\sqrt{2}}$

Answer 2

Answer:

The time taken to cover the first half of the height is $\dfrac{t_{1}}{\sqrt{2}}$.

Explanation:

Given that,

Height = h

Time = t₁

We need to calculate the time ,

Using equation of motion

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

$h=\dfrac{1}{2}gt_{1}^2$

The time taken to cover the first half of the height is

$s=\dfrac{h}{2}$

Using equation of motion

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

Put the value into the formula

$\dfrac{h}{2}=\dfrac{1}{2}gt^2$....(II)

Put the value of h in the equation (II)

$\dfrac{\dfrac{1}{2}gt_{1}^2}{2}=\dfrac{1}{2}gt^2$

$t=\dfrac{t_{1}}{\sqrt{2}}$

Hence, The time taken to cover the first half of the height is $\dfrac{t_{1}}{\sqrt{2}}$.