A body is released from top of a tower of height h . It takes t sec to reach ground. Where will be b

English

Gujarati

Hindi

2.Motion in Straight Line

medium

A body is released from the top of a tower of height $h$. It takes $t$ sec to reach the ground. Where will be the ball after time $t/2$ sec

A

At $h/2$from the ground

B

At $h/4$ from the ground

C

Depends upon mass and volume of the body

D

At $3h/4$ from the ground

Solution

(d) Let the body after time $t/2$ be at $x$ from the top, then

$x = \frac{1}{2}g\frac{{{t^2}}}{4} = \frac{{g{t^2}}}{8}$…(i)

$h = \frac{1}{2}g{t^2}$…(ii)

Eliminate $t$ from (i) and (ii), we get $x = \frac{h}{4}$

$\therefore $ Height of the body from the ground $ = h – \frac{h}{4} = \frac{{3h}}{4}$

Standard 11

Physics

Share

Similar Questions

Velocity of a body on reaching the point from which it was projected upwards, is

easy

(AIIMS-1999)

A body falling under gravity covers two points ${A}$ and ${B}$ separated by $80\ m$ in $2 \ s$. The distance of upper point $A$ from the starting point is ${m}$ (use ${g}=10 \ m s^{-2}$).

hard

(JEE MAIN-2024)

A balloon is at a height of $81\, m$ and is ascending upwards with a velocity of $12 \,m/s$. A body of $2\,kg$ weight is dropped from it. If $g = 10\,m/{s^2}$, the body will reach the surface of the earth in………$s$

medium

A ball of mass $0.5 \; kg$ is dropped from the height of $10 \; m$. The height, at which the magnitude of velocity becomes equal to the magnitude of acceleration due to gravity, is $\dots \; m$. (Use $g =10 \; m / s ^{2}$).

medium

(JEE MAIN-2022)

Three different objects of masses ${m_1},{m_2}$ and ${m_3}$ are allowed to fall from rest and from the same point $‘O’$ along three different frictionless paths. The speeds of the three objects, on reaching the ground, will be in the ratio of

easy

(AIIMS-2002)