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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 $g = 10 m/s^{2}$)
Explanation
Clearly distance moved by $1^{st}$ ball in 18 s = distance moved by $2^{nd}$ ball in 12 s. Now, distance moved in 18 s by $1^{st}$ ball = $\frac{1}{2} \times 10 \times 18^{2} = 90 \times 18 = 1620 m$ Distance moved in 12 s by $2^{nd}$ ball = $ut + \frac{1}{2}gtt^{2} \therefore 1620 = 12v + 5 \times 144$ $\Rightarrow v = 135 - 60 = 75 ms^{-1}$.