A ball is dropped from a high rise platform at $t=0$ starting from rest. After 4 seconds another bal

Question

A ball is dropped from a high rise platform at $t=0$ starting from rest. After 4 seconds another ball is thrown downwards from the same platform with a speed $v$. The two balls meet at $t=12 \mathrm{~s}$. What is the value of $v$? (Take $g=10 \mathrm{~m} / \mathrm{s}^{2}$)

Professor PC Thomas Classes · Answer

(b) : Let the two balls meet after $12 \mathrm{~s}$ at distance $x$ from the platform. \ For the first ball, $u=0, t=12 \mathrm{~s}, g=10 \mathrm{~m} / \mathrm{s}^{2}$ \ $	herefore \quad x=\frac{1}{2} 	imes 10 	imes 12^{2} = 5 	imes 144 = 720 \mathrm{~m}$ \ For the second ball, $u=v, t=(12-4)=8 \mathrm{~s}, g=10 \mathrm{~m} / \mathrm{s}^{2}$ \ $	herefore \quad x=v 	imes 8+\frac{1}{2} 	imes 10 	imes 8^{2} = 8v + 320 \mathrm{~m}$ \ Equating the distances: $720 = 8v + 320 \implies 400 = 8v \implies v=50 \mathrm{~m} / \mathrm{s}$