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JEE Main 2021 (26 Aug, Shift 1) · Motion Under Gravity

Motion Under Gravity question from JEE Main 2021 (26 Aug, Shift 1)

Q36JEE Main 2021 · 26 AUG · SHIFT 1Medium

Motion Under Gravity

Two spherical balls having equal masses with radius of 5 cm each are thrown upwards along the same vertical direction at an interval of 3s with the same initial velocity of 35 m/s, then these balls collide at a height of ............... m. (Take g = 10 m/s2)

Numerical answer type. Work it out, then open the answer.

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Answer: 50

  1. y1=35t−5t2y_1 = 35t - 5t^2, y2=35(t−3)−5(t−3)2y_2 = 35(t-3) - 5(t-3)^2
  2. Equal heights: 0=−105+30t−45⇒t=5 s0 = -105 + 30t - 45 \Rightarrow t = 5\text{ s}
  3. Height =35(5)−5(5)2=175−125=50 m= 35(5) - 5(5)^2 = 175 - 125 = 50\text{ m} (the 5 cm radius is negligible).
  4. Collision height =50 m= 50\text{ m}
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