A particle is released from rest from a tower | vedclass

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

Question

(D) Using the equation of motion $S = ut + \frac{1}{2}at^2$,where $u = 0$ and $a = g$,we get $S = \frac{1}{2}gt^2$,which implies $t = \sqrt{\frac{2S}{

Vedclass · Accepted Answer

$1 : (\sqrt{2} - 1) : (\sqrt{3} - \sqrt{2})$

Vedclass · Answer

(D) Using the equation of motion $S = ut + \frac{1}{2}at^2$,where $u = 0$ and $a = g$,we get $S = \frac{1}{2}gt^2$,which implies $t = \sqrt{\frac{2S}{g}}$.For the first height $h$,the time taken is $t_1 = \sqrt{\frac{2h}{g}}$.For the first two heights (total distance $2h$),the time taken is $T_2 = \sqrt{\frac{2(2h)}{g}} = \sqrt{\frac{4h}{g}}$. Thus,the time taken for the second interval $h$ is $t_2 = T_2 - t_1 = \sqrt{\frac{4h}{g}} - \sqrt{\frac{2h}{g}} = \sqrt{\frac{2h}{g}}(\sqrt{2} - 1)$.For the first three heights (total distance $3h$),the time taken is $T_3 = \sqrt{\frac{2(3h)}{g}} = \sqrt{\frac{6h}{g}}$. Thus,the time taken for the third interval $h$ is $t_3 = T_3 - T_2 = \sqrt{\frac{6h}{g}} - \sqrt{\frac{4h}{g}} = \sqrt{\frac{2h}{g}}(\sqrt{3} - \sqrt{2})$.Taking the ratio $t_1 : t_2 : t_3$:$t_1 : t_2 : t_3 = \sqrt{\frac{2h}{g}} : \sqrt{\frac{2h}{g}}(\sqrt{2} - 1) : \sqrt{\frac{2h}{g}}(\sqrt{3} - \sqrt{2})$$t_1 : t_2 : t_3 = 1 : (\sqrt{2} - 1) : (\sqrt{3} - \sqrt{2})$.

Column $I$	Column $II$
$(A)$ Height of particle from ground	$(p)$ $0$
$(B)$ Speed of particle	$(q)$ $10$ $SI$ unit
$(C)$ Displacement of particle	$(r)$ $40$ $SI$ unit
$(D)$ Acceleration of particle	$(s)$ $20$ $SI$ unit

$A$ particle is released from rest from a tower of height $3h$. The ratio of times taken to fall through successive equal heights $h$,i.e.,$t_1 : t_2 : t_3$ is

Similar Questions

$A$ ball is dropped from a bridge that is $45 \ m$ above the water. It falls directly into a boat which is moving with constant velocity. The boat is $12 \ m$ away from the point of impact when the ball is dropped. The speed of the boat is (Take $g=10 \ m \ s^{-2}$) (in $m \ s^{-1}$)

$A$ juggler tosses a ball up in the air with initial speed $u$. At the instant it reaches its maximum height $H$,he tosses up a second ball with the same initial speed. The two balls will collide at a height:

$A$ ball is dropped from a height $h$ above the ground. Neglecting air resistance,its velocity $(v)$ varies with its height $(y)$ above the ground as:

Vedclass Test Series

Exam Paper Generator

Online Exam Module