Question

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

Answer 1

Answer:$v_0=\sqrt{gH}$

Explanation:

Given

body A is from a height H and simultaneously body B is thrown upward with a velocity $v_0$ and they both meet at $\frac{H}{2}$

For body A

$\frac{H}{2}=ut+\frac{at^2}{2}$

$\frac{H}{2}=0+\frac{gt^2}{2}$-----1

For Body B

$\frac{H}{2}=v_0t-\frac{gt^2}{2}$-------2

$t=\sqrt{\frac{H}{g}}$

put t in 2

$\frac{H}{2}=v_0\times \sqrt{\frac{H}{g}}-\frac{g}{2}\times \frac{H}{g}$

$v_0=\sqrt{gH}$

Answer 2

Given:

Initial speed of body A,u=0

Initial Speed of body B=v0

Height from which body is doped=H

They meet at height=H/2

To find:

The value of v0

Solution:

We know that

$S=ut+1/2 gt^2$

Using the formula

For body A

$H/2=ut+\frac{1}{2}gt^2$

$H/2=0+1/2gt^2$

$H/2=1/2gt^2$

$H=gt^2$

$t^2=H/g$

$t=\sqrt{H/g}$

For body B

$H/2=v_0t-1/2gt^2$

Substitute the value of t

$H/2=v_0\times \sqrt{H/g}-\frac{1}{2}g\times H/g$

$H/2=\sqrt{H/g}v_0-1/2H$

$H/2+1/2H=\sqrt{H/g}v_0$

$H=\sqrt{H/g}v_0$

$H\times \sqrt{g/H}=v_0$

$\sqrt{gH}=v_0$

Hence, they meet at height =H/2 if v0 =$\sqrt{gH}$