#### A ball is dropped from a high-rise platform at $\mathrm{t}=0$  starting from rest. After 6 seconds another ball is thrown downwards from the same platform with a speed  $v$. The two balls meet at  $t=18 \mathrm{~s}$. What is the value of $v$ ? (Take $g=10 \mathrm{~m} / \mathrm{s}^2$ )Option: 1 $75\ \frac{m}{s}$Option: 2 $80\ \frac{m}{s}$Option: 3 $120\ \frac{m}{s}$Option: 4 None of the above.

For ball 1:

$h=\frac{1}{2}gt^2$

$= \frac{1}{2}\times10\times{18}^2=1620\ m$

For ball 2:

$h=1620m,\ u=v,t=18-6=12s$

Therefore,  $h=ut+\frac{1}{2}gt^2$

Or, $1620=12v+\frac{1}{2}\times10\times{12}^2$

Or, $12v=900$

Or, $v=75\ \frac{m}{s}$
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