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)

$75\,m/s$

Answer 2

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 \hspace30mm ...(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 \hspace15mm ...(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 $\, \, v = \frac{1 \times 10 \times 30 \times 6 }{2 \times12}=75\, m/s$