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 $t = 18 \, s$. What is the value of $v$?
(Take $g = 10\, m/s^2$ )

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

Answer 1

Answer: (A)

Let the two balls meet after $t \,s$ at distance $x$ from the platform.
For the first ball
$u = 0, t = 18\, s, g = 10\, m/s^2$
Using $ h = ut+\frac{1}{2} gt^2$
$\therefore x = \frac{1}{2} \times 10 \times 18^2$...(i)
For the second ball $ u = v, t = 12\, s, g = 10\, m/s^2$
Using $h = ut+\frac{1}{2} gt^2$
$\therefore x = v \times 12 + \frac{1}{2} \times 10 \times 12^2$...(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 $v = \frac{1 \times 10 \times 30 \times 6 }{2 \times12}=75\, m/s$