A stone is thrown vertically upward. On its way up it passes point A with speed of v, and point B, 3 m highter than A, with speed V / 2 The maximum height reached by stone above point B is :-

Question

A stone is thrown vertically upward. On its way up it passes point A with speed of v, and point B, 3 m highter than A, with speed V / 2 The maximum height reached by stone above point B is :- (A) 1 m (B) 2 m (C) 3 m (D) 5 m

Tardigrade Admin · Accepted Answer

((v/2))2=v2-2 g × 3 ∴ v=√8 g If h is the further height, then 0=((v/2))2-2 g h ∴ h=(v2/8 g)=(8 g/8 g)=1 m

Q. A stone is thrown vertically upward. On its way up it passes point $A$ with speed of $v$ , and point $B, 3 m$ highter than $A$ , with speed $V /2$ The maximum height reached by stone above point $B$ is :-

1 m

2 m

3 m

5 m

$(\frac{v}{2})^{2} = v^{2} - 2 g \times 3$
$∴ v = 8 g$
If $h$ is the further height, then
$0 = (\frac{v}{2})^{2} - 2 g h$
$∴ h = \frac{v ^{2}}{8 g} = \frac{8 g}{8 g} = 1 m$

Q. A stone is thrown vertically upward. On its way up it passes point A with speed of v, and point B,3m highter than A, with speed V/2 The maximum height reached by stone above point B is :-

1 m

2 m

3 m

5 m

(2v​)2=v2−2g×3 ∴v=8g​ If h is the further height, then 0=(2v​)2−2gh ∴h=8gv2​=8g8g​=1m

Q. A stone is thrown vertically upward. On its way up it passes point $A$ with speed of $v$ , and point $B, 3 m$ highter than $A$ , with speed $V /2$ The maximum height reached by stone above point $B$ is :-

$(\frac{v}{2})^{2} = v^{2} - 2 g \times 3$
$∴ v = 8 g$
If $h$ is the further height, then
$0 = (\frac{v}{2})^{2} - 2 g h$
$∴ h = \frac{v ^{2}}{8 g} = \frac{8 g}{8 g} = 1 m$