48. 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/s2 )

48. 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 :  (a) 75 m/s

Taking downward direction as positive, 

\( \text{Ball A travels for 18 s to reach } S_A \text{ where both balls are at equal height} \)

\( S_A = ut + \)\( \frac{1}{2} \)\( at^2 \)

\( = 0 + \)\( \frac{1}{2} \)\( (10)(18^2) \)

\( \text{Ball B travels for } (18 - 6) = 12 \text{ s to reach } S_B \text{ where both balls are at equal height} \)

\( S_B = vt + \)\( \frac{1}{2} \)\( at^2 \)

\( = v(12) + \)\( \frac{1}{2} \)\( (10)(12^2) \)

\( S_A = S_B \)

\( \rightarrow \)\( \frac{1}{2} \)\( 10(18^2) = v(12) + \)\( \frac{1}{2} \)\( 10(12^2) \)

\( \rightarrow5 \times 180 = 12v \)

\( \rightarrow v = \)\( \frac{5 \times 180}{12} \)\( \)

\( = 75 \, \text{m/s} \)