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. Taking g=10 m/s², the value of v is

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(1) 75 m/s. Given: first ball dropped at t=0; second thrown downward at t=6 s with speed v; they meet at t=18 s with g=10 m/s². Idea: at the meeting instant both balls have fallen the same distance below the platform, but for different lengths of time. First ball, falling for 18 s from rest: x=1/2(10)(18)²=1620 m. Second ball, falling for 18-6=12 s with initial speed v: x=12v+1/2(10)(12)²=12v+720. Equating: 1620=12v+720. 12v=900, so v=75 m/s.

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. Taking g=10 m/s², the value of v is

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