Question

Object $A$ is dropped from a height $h$. At the same instant object-$B$ is thrown vertically upward from the ground. Right before they collide in mid-air, the speed of $A$ is twice the speed of $B$. Determine the height where the collision occurs.

• A. $\frac{2h}{3}$
• B. $\frac{h}{\sqrt{3}}$
• C. $\frac{3h}{4}$
• D. $\frac{h}{2}$

Answer 1

The correct option is A $\frac{2h}{3}$

Let $(1-x)$ be the height for A and $x$ be the height for B

For A,

$(1-x)h=\frac{1}{2}g{t}^2$

$\therefore t=\sqrt{\frac{2(1-x)h}{g}}$

For B,

$xh=ut-\frac{1}{2}g{t}^2$

$\Rightarrow xh=u\left(\sqrt{\frac{2(1-x)h}{g}}\right)-(1-x)h$

solving, we get,

$u=\sqrt{\frac{gh}{2(1-x)}}$

$v_A=2v_B$

$2g(1-x)h=4(u^2-2ghx)$

$\therefore x=\frac{2}{3}h$