A body is dropped from a height H. The time t | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let $T$ be the total time taken to fall from height $H$. Using the equation of motion $s = ut + \frac{1}{2}gt^2$ with initial velocity $u = 0$:$H

Vedclass · Accepted Answer

$\sqrt{\frac{H}{g}}(\sqrt{2}-1)$

Vedclass · Answer

(C) Let $T$ be the total time taken to fall from height $H$. Using the equation of motion $s = ut + \frac{1}{2}gt^2$ with initial velocity $u = 0$:$H = \frac{1}{2} g T^2 \Rightarrow T = \sqrt{\frac{2H}{g}}$Let $t_1$ be the time taken to cover the first half of the journey,i.e.,distance $H/2$:$\frac{H}{2} = \frac{1}{2} g t_1^2 \Rightarrow t_1 = \sqrt{\frac{H}{g}}$The time taken to cover the second half of the journey is the difference between the total time and the time taken for the first half:$t_2 = T - t_1 = \sqrt{\frac{2H}{g}} - \sqrt{\frac{H}{g}}$$t_2 = \sqrt{\frac{H}{g}} (\sqrt{2} - 1)$

$A$ body is dropped from a height $H$. The time taken to cover the second half of the journey is:

Similar Questions

An object is projected upwards from the foot of a tower. The object crosses the top of the tower twice with an interval of $8 \ s$ and the object reaches the foot again after $16 \ s$. The height of the tower is [ $g = 10 \ m/s^2$ ]. (in $m$)

$A$ ball is projected upwards. Its acceleration at the highest point is

$A$ boy standing at the top of a tower of $20 \, m$ height drops a stone. Assuming $g = 10 \, m/s^2$,the velocity with which it hits the ground is ......... $m/s$.

$A$ ball is thrown vertically upward. It has a speed of $10 \; m/s$ when it has reached one-half of its maximum height. How high does the ball rise? (Take $g = 10 \; m/s^2$)

The effective acceleration of a body,when thrown upwards with acceleration $a$,will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module