A stone falls freely under gravity. It covers | vedclass

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

Question

(B) For a body falling freely from rest $(u = 0)$,the distance covered in time $t$ is given by $h = \frac{1}{2}gt^2$.$1$. Distance covered in the firs

Vedclass · Accepted Answer

$h_1 = \frac{h_2}{3} = \frac{h_3}{5}$

Vedclass · Answer

(B) For a body falling freely from rest $(u = 0)$,the distance covered in time $t$ is given by $h = \frac{1}{2}gt^2$.$1$. Distance covered in the first $5 \ s$ $(t = 5 \ s)$:$h_1 = \frac{1}{2}g(5)^2 = \frac{25}{2}g$ ... $(i)$$2$. Distance covered in the first $10 \ s$ $(t = 10 \ s)$:$h_1 + h_2 = \frac{1}{2}g(10)^2 = \frac{100}{2}g$ ... (ii)$3$. Distance covered in the first $15 \ s$ $(t = 15 \ s)$:$h_1 + h_2 + h_3 = \frac{1}{2}g(15)^2 = \frac{225}{2}g$ ... (iii)Subtracting $(i)$ from (ii):$h_2 = (h_1 + h_2) - h_1 = \frac{100}{2}g - \frac{25}{2}g = \frac{75}{2}g = 3 \left( \frac{25}{2}g \right) = 3h_1$Subtracting (ii) from (iii):$h_3 = (h_1 + h_2 + h_3) - (h_1 + h_2) = \frac{225}{2}g - \frac{100}{2}g = \frac{125}{2}g = 5 \left( \frac{25}{2}g \right) = 5h_1$Thus,$h_1 = \frac{h_2}{3} = \frac{h_3}{5}$.

$A$ stone falls freely under gravity. It covers distances $h_1, h_2$ and $h_3$ in the first $5 \ s$,the next $5 \ s$ and the next $5 \ s$ respectively. The relation between $h_1, h_2$ and $h_3$ is:

Similar Questions

$A$ stone is allowed to fall from the top of a tower $100 \ m$ high and at the same time another stone is projected vertically upwards from the ground with a velocity of $25 \ m/s$. The two stones will meet after ......... $s$.

From a tower of height $H$, a particle is thrown vertically upwards with a speed $u$. The time taken by the particle to hit the ground is $n$ times that taken by it to reach the highest point of its path. The relation between $H, u$ and $n$ is:

$A$ person throws balls into the air vertically upward at regular intervals of time of $1 \,s$. The next ball is thrown when the velocity of the ball thrown earlier becomes zero. The height to which the balls rise is (Assume $g = 10 \,m/s^2$) (in $\,m$)

$A$ ball is projected vertically upwards from the ground. It reaches a height $h$ in time $t_1$,continues its motion,and then takes a time $t_2$ to reach the ground. The height $h$ in terms of $g, t_1$,and $t_2$ is ($g =$ acceleration due to gravity).

$A$ particle is dropped under gravity from rest from a height $h$ $(g = 9.8 \ m/s^2)$. If it travels a distance of $9h/25$ in the last second of its motion,the height $h$ is ......... $m$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module