Question

Ball A is dropped freely while another ball B is thrown vertically downward with an initial velocity $v$ from the same point simultaneously. After $tseconds$ they are separated by a distance of: (Given that $t$ is less than the time required by ball B to reach the ground.)

• A. $\frac{vt}{2}$
• B. $\frac{1}{2}g{t}^2$
• C. $vt$
• D. $vt+\frac{1}{2}g{t}^2$

Answer 1

Answer: (C)

$vt$

First case (for ball A):
$s_1=0\left(t\right)+\frac{1}{2}g{t}^2$
Second case (for ball B):
$s_2=v\left(t\right)+\frac{1}{2}g{t}^2$
after $tseconds$, distance between them is $s_2-s_1$
$\Rightarrow s_2-s_1=vt$

Note: $\text{After ball B has landed on ground the distance between A and B decreases and becomes zero.}$

So, C is correct.