Question

Two balls $A$ and $B$ are placed at the top of $180 \,m$ tall tower. Ball $A$ is released from the top at $t =0\, s$. Ball $B$ is thrown vertically down with an initial velocity ' $u$ ' at $t =2\, s$. After a certain time, both balls meet $100 \,m$ above the ground. Find the value of '$u$' in $ms ^{-1}$. [use $g =10 \,ms ^{-2}$ ] :

• A. 10
• B. 15
• C. 20
• D. 30

Answer 1

Answer: (D)

Let they meet at time $t$.
$t=\sqrt{\frac{2 h}{g}}=\sqrt{\frac{2 \times 80}{10}} $
$=4\, \sec$
Time taken by ball $B$ to meet $A =2 \,\sec$
using $ S=u t+\frac{1}{2} at ^{2}$
$-80=- u \times 2+\frac{1}{2}(-10)(2)^{2} $
$ u =30$