A ball is dropped freely While another is thrown vertically downward with an Initial velocity v from the same point simultaneously. After 't' second they are separated by a distance of

Question

A ball is dropped freely While another is thrown vertically downward with an Initial velocity v from the same point simultaneously. After 't' second they are separated by a distance of (A) (Vt/2) (B) (1/2)gt2 (C) vt (D) ( vt + gt 2/2)

Tardigrade Admin · Accepted Answer

ball 1: u =0 v = u + at v = gt S 1=( ut + at 2/2) S 1=( gt 2/2) ball 2: u = v vf = u + at vf = v + gt S 2=( vt + gt 2/2) distance of separation = S 2- S 1= vt.

Q. A ball is dropped freely While another is thrown vertically downward with an Initial velocity $v$ from the same point simultaneously. After $^{'} t^{'}$ second they are separated by a distance of

$\frac{V t}{2}$

$\frac{1}{2} g t^{2}$

vt

$\frac{v t + g t ^{2}}{2}$

ball 1:
$u = 0$
$v = u + a t$
$v = g t$
$S 1 = \frac{u t + a t ^{2}}{2}$
$S 1 = \frac{g t ^{2}}{2}$
ball 2:
$u = v$
$v f = u + a t$
$v f = v + g t$
$S 2 = \frac{v t + g t ^{2}}{2}$
distance of separation $= S 2 - S 1 = v t$ .

Q. A ball is dropped freely While another is thrown vertically downward with an Initial velocity v from the same point simultaneously. After ′t′ second they are separated by a distance of

2Vt​

21​gt2

vt

2vt+gt2​

ball 1: u=0 v=u+at v=gt S1=2ut+at2​ S1=2gt2​ ball 2: u=v vf=u+at vf=v+gt S2=2vt+gt2​ distance of separation =S2−S1=vt.

Q. A ball is dropped freely While another is thrown vertically downward with an Initial velocity $v$ from the same point simultaneously. After $^{'} t^{'}$ second they are separated by a distance of

$\frac{V t}{2}$

$\frac{1}{2} g t^{2}$

$\frac{v t + g t ^{2}}{2}$

ball 1:
$u = 0$
$v = u + a t$
$v = g t$
$S 1 = \frac{u t + a t ^{2}}{2}$
$S 1 = \frac{g t ^{2}}{2}$
ball 2:
$u = v$
$v f = u + a t$
$v f = v + g t$
$S 2 = \frac{v t + g t ^{2}}{2}$
distance of separation $= S 2 - S 1 = v t$ .