A ball is thrown vertically upwards from the top of a tower of height h with velocity v. The ball strikes the ground after

Question

A ball is thrown vertically upwards from the top of a tower of height h with velocity v. The ball strikes the ground after (A) (v/g)[1+√1+(2 g h/v2)] (B) (v/g)[1+√1-(2 g h/v2)] (C) (v/g)(1+(2 g h/v2))1 / 2 (D) (v/g)(1-(2 g h/v2))1 / 2

Tardigrade Admin · Accepted Answer

As, h=-v t+(1/2) g t2 or f t2-2 v t-h=0 ⇒ t=(-(-2 v) ± √4 v2+4 g h/2 g)=(2 v ± √v2+g h/2 g) =(v/g) ± ([v2+2 g h]1 / 2/g)=(v/g)[1 ± √1+(2 g h/v2)]

Q. A ball is thrown vertically upwards from the top of a tower of height $h$ with velocity $v$ . The ball strikes the ground after

$\frac{v}{g} [1 + 1 + \frac{2 g h}{v ^{2}}]$

$\frac{v}{g} [1 + 1 - \frac{2 g h}{v ^{2}}]$

$\frac{v}{g} (1 + \frac{2 g h}{v ^{2}})^{1/2}$

$\frac{v}{g} (1 - \frac{2 g h}{v ^{2}})^{1/2}$

As, $h = - v t + \frac{1}{2} g t^{2}$
or $f t^{2} - 2 v t - h = 0$
$\Rightarrow t = \frac{- ( - 2 v ) \pm 4 v ^{2} + 4 g h}{2 g} = \frac{2 v \pm v ^{2} + g h}{2 g}$
$= \frac{v}{g} \pm \frac{[ v ^{2} + 2 g h ] ^{1/2}}{g} = \frac{v}{g} [1 \pm 1 + \frac{2 g h}{v ^{2}}]$

Q. A ball is thrown vertically upwards from the top of a tower of height h with velocity v. The ball strikes the ground after

gv​[1+1+v22gh​​]

gv​[1+1−v22gh​​]

gv​(1+v22gh​)1/2

gv​(1−v22gh​)1/2