JEE Main 2022 · 29 Jun · Shift 1 · Physics Q31

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PhysicsSingle correct+4 / -1Question 31 of 90

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 $^{-2}$ ] :

Answer and worked solution

Official answer: D

Ball A falls 80 m: $t=4$ s. Ball B has 2 s: $80=2u+\frac12\cdot10\cdot4$ , so $u=30$ m s $^{-1}$ .

(D)

Topics:Motion in a straight line

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JEE Main 2022 · 29 Jun · Shift 1, Physics question 31