Byju's Answer

Standard XII

Physics

Dimensional Analysis

A body falls ...

Question

A body falls freely under gravity from rest and reaches the ground in time $t$ . Write expression for the height fallen by the body.

H = \frac{1}{2}

g t^{2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

H = u t + \frac{1}{2}

g t^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

H = \frac{1}{2}

g t

No worries! We‘ve got your back. Try BYJU‘S free classes today!

H = u t - \frac{1}{2}

g t^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $H = \frac{1}{2}$ $g t^{2}$
We know that second equation of motion is given by
$S = u t + \frac{1}{2} g t^{2}$
where $S = distance, u = initial velocity = 0 , t = time , g = acc. due to gravity$
$\Rightarrow S = \frac{1}{2} g t^{2}$
Since, $u = 0$

Suggest Corrections

Similar questions

A body is dropped from a certain height at time $T = 0$ . It undergoes freefall and reaches the ground at time $T = t$ s. What is the expression for the height $h$ from which the body started its free fall?

Acceleration due to gravity is $g$ .

Q. Prove with the help of dimensional analysis that the equation

h = \frac{1}{2} g t

for the distance travelled by a body falling freely under the gravity in time t is incorrect. Find the correct equation with the help of dimensions.

Q. If a ball fallen freely from 'h' height reaches in time 't' at ground, then what will be the time when it reaches at height

\frac{h}{2}

Q. The height

h

reached by a particle thrown up vertically as a function of times is given by

h = u t - \frac{1}{2} g t^{2}

where

u

and

g

are constants. The maximum height is

Q. A stone falls freely under gravity. It covers distances

h_{1}, h_{2}

and

h_{3}

in the first

5 s e c o n d

, the next

5 s

and next

5 s e c o n d

respectively. The relation between

h_{1}, h_{2}

and

h_{3}

is :

Join BYJU'S Learning Program

Select...

A body falls freely under gravity from rest and reaches the ground in time t. Write expression for the height fallen by the body.

The correct option is A H=12 gt2We know that second equation of motion is given byS=ut+12gt2where S = distance, u = initial velocity = 0 , t = time , g = acc. due to gravity⇒S=12gt2Since, u=0

A body falls freely under gravity from rest and reaches the ground in time $t$ . Write expression for the height fallen by the body.

The correct option is A $H = \frac{1}{2}$ $g t^{2}$
We know that second equation of motion is given by
$S = u t + \frac{1}{2} g t^{2}$
where $S = distance, u = initial velocity = 0 , t = time , g = acc. due to gravity$
$\Rightarrow S = \frac{1}{2} g t^{2}$
Since, $u = 0$