Question

A ball is dropped from a high rise platform at $t=0$ starting from rest. After $6$ seconds another ball is thrown downwards from the same platform with a speed $v$. The two balls meet at $t=18s$. What is the value of $v$?
(Take $g=10m/s^2$ )

• A. $75m/s$
• B. $55m/s$
• C. $40m/s$
• D. $60m/s$

Answer 1

Answer: (A)

Let the two balls meet after $ts$ at distance $x$ from the platform.
For the first ball
$u=0,t=18s,g=10m/s^2$
Using $h=ut+\frac{1}{2}g{t}^2$
$\therefore x=\frac{1}{2}\times 10\times 18^2$...(i)
For the second ball $u=v,t=12s,g=10m/s^2$
Using $h=ut+\frac{1}{2}g{t}^2$
$\therefore x=v\times 12+\frac{1}{2}\times 10\times 12^2$...(ii)
From equations (i) and (ii), we get
$\frac{1}{2}\times 10\times 18^2=12v+\frac{1}{2}\times 10\times (12{)}^2$
or $12v=\frac{1}{2}\times 10\times [(18{)}^2-(12{)}^2]$
$=\frac{1}{2}\times 10\times [(18+12)-(18-12)]$
$12v=\frac{1}{2}\times 10\times 30\times 6$
or $v=\frac{1\times 10\times 30\times 6}{2\times 12}=75m/s$