The ratio of the displacements of a freely fa | vedclass

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

Question

(D) The displacement of a body in the $n^{th}$ second is given by the formula: $S_n = u + \frac{a}{2}(2n - 1)$.For a freely falling body,the initial v

Vedclass · Accepted Answer

$3: 9$

Vedclass · Answer

(D) The displacement of a body in the $n^{th}$ second is given by the formula: $S_n = u + \frac{a}{2}(2n - 1)$.For a freely falling body,the initial velocity $u = 0$ and acceleration $a = g$.Thus,the displacement in the $n^{th}$ second is $S_n = \frac{g}{2}(2n - 1)$.For the second second $(n = 2)$: $S_2 = \frac{g}{2}(2(2) - 1) = \frac{g}{2}(3) = 1.5g$.For the fifth second $(n = 5)$: $S_5 = \frac{g}{2}(2(5) - 1) = \frac{g}{2}(9) = 4.5g$.The ratio of displacements is $\frac{S_2}{S_5} = \frac{1.5g}{4.5g} = \frac{1.5}{4.5} = \frac{1}{3}$.

The ratio of the displacements of a freely falling body during the second and fifth seconds of its motion is

Similar Questions

$A$ 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 acceleration of a vertically projected body at its highest reaching position is

$A$ stone thrown upward with a speed $u$ from the top of a tower reaches the ground with a velocity $3u$. The height of the tower is (in $u^2/g$)

$A$ body dropped from a height $h$ with an initial speed zero,strikes the ground with a velocity $3 \ km/h$. Another body of the same mass is dropped from the same height $h$ with an initial speed of $4 \ km/h$. Find the final velocity of the second body with which it strikes the ground in $km/h$.

$Assertion$ : $A$ body can have acceleration even if its velocity is zero at a given instant of time.
$Reason$ : $A$ body is momentarily at rest when it reverses its direction.

Vedclass Test Series

Exam Paper Generator

Online Exam Module