Question

A body ‘ $A$’ is dropped vertically from the top of a tower. If another identical body ‘$B$’ is projected from the same point at the same instant, then

• A. both ‘A’ and ‘B’ will reach the ground simultaneously
• B. ‘A’ will reach the ground earlier than ‘B’
• C. ‘B’ will reach the ground earlier than ‘A’
• D. either ‘A’ or ‘B’

Answer 1

Answer: (A)

As the body ' $A$ ' is dropped from rest
$\therefore h=\frac{1}{2}g{t}_A^2$
$\Rightarrow t_A=\sqrt{\frac{2h}{g}}$

As the body $B$ is given a horizontal velocity at the time of release, it is going to follow the same trajectory as a body on a projectile motion projected with a velocity having the same horizontal component as the horizontal velocity of $B$ given at $H$. So the time taken by $B$ from $H$ to $Y$ is same as that from $X$ to $H$.
Now let the vertical component of velocity at $X$ be $v$ then at $H$
$0=v^2-2gh\left[\therefore atH,v_f=0\right]$
$\Rightarrow v^2=2gh$
$\Rightarrow v=\sqrt{2gh}$
and $0=v-gt$
$\Rightarrow t=\frac{v}{g}=\frac{\sqrt{2gh}}{g}=\sqrt{\frac{2h}{g}}$
$\therefore t_B=t=\sqrt{\frac{2h}{g}}=t_A$