Question

A ball is dropped freely While another is thrown vertically downward with an Initial velocity $v$ from the same point simultaneously. After $'t'$ second they are separated by a distance of

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

Answer 1

Answer: (C)

ball 1:
$u =0$
$v = u + at$
$v = gt$
$S 1=\frac{ ut + at ^{2}}{2}$
$S 1=\frac{ gt ^{2}}{2}$
ball 2:
$u = v$
$vf = u + at$
$vf = v + gt$
$S 2=\frac{ vt + gt ^{2}}{2}$
distance of separation $= S 2- S 1= vt$.