A body A is thrown up vertically from the ground with a velocity v0 and another body B is simultaneously dropped from a height H. They meet at a height (H/2) if v0 is equal to

Question

A body A is thrown up vertically from the ground with a velocity v0 and another body B is simultaneously dropped from a height H. They meet at a height (H/2) if v0 is equal to (A)  √2gH  (B)  √gH  (C)  (1/2)√gH  (D)  (2g/H)

Tardigrade Admin · Accepted Answer

Suppose the two bodies A and B meet at time t, at a height (H/2) from the ground. For body B, u=0, h=(H/2) h=u t+(1/2) g t2 (H/2)=(1/2) g t2 ...(i) For body A, d=v0, h=(H/2) h =u t-(1/2) g t2 (H/2) =v0 t-(1/2) g t2 ...(ii) From Eqs. (i) and (ii) v0 t-(1/2) g t2=(1/2) g t2 or v0 t=g t2 or t=(v0/g) Substituting the value of t in Eq. (i), we get (H/2) =(1/2) g t2 H =g ×((v0/g))2 v0 =√g H

Q. A body $A$ is thrown up vertically from the ground with a velocity $v_{0}$ and another body $B$ is simultaneously dropped from a height $H$ . They meet at a height $\frac{H}{2}$ if $v_{0}$ is equal to

$2 g H$

$g H$

$\frac{1}{2} g H$

$\frac{2 g}{H}$

Q. A body A is thrown up vertically from the ground with a velocity v0​ and another body B is simultaneously dropped from a height H. They meet at a height 2H​ if v0​ is equal to

2gH​

gH​

21​gH​

H2g​

Q. A body $A$ is thrown up vertically from the ground with a velocity $v_{0}$ and another body $B$ is simultaneously dropped from a height $H$ . They meet at a height $\frac{H}{2}$ if $v_{0}$ is equal to

$2 g H$

$g H$

$\frac{1}{2} g H$

$\frac{2 g}{H}$