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`        a man drops a ball downside from teh roof of a tower of hight 400 meters. At same time another ball is thrown upside with a velocity 50m/s from teh base of the tower,then they will meet at which height from the base of the tower.`
one month ago

sumit hajare
28 Points
```							A ball dropped from the roof of height 400m Let after t time it meets with another ball .displacement covered by ball during this time ( S1 ) = ut + 1/2at² = 0× t + 1/2 × 10 × t² = 5t² another ball is thrown from the surface of tower with speed 50m/sec .then, displacement covered by another ball (S2) = ut +1/2at² = 50t - 1/2× 10t² =50t -5t² for collision, S1 + S2 = 400 5t² + 50t - 5t² = 400 50t = 400 t = 8sec hence, body collide in 8sec.then, displacement covered by first body during this time= 5 × 64 = 320m e.g collision occurs 80m height from the surface of tower
```
one month ago
Vikas TU
8825 Points
```							Dear student A ball dropped from the roof of height 400m Let after t time it meets with another ballSo,400-x=50_/(x/5) - 5(_/(x/5))^2=>400-x=10_/(5x) - x=>400=10_/(5x)SQUARING BOTH THE SIDES=>160000=500x=>x=1600/5=320 mSo, height of collision from ground or surface of tower is (400-320)=80 m
```
one month ago
Nitin Kumar
26 Points
```							You can simplify time calculation by using relative motion of the two balls. As seen from the top ball. The bottom ball is moving at a relative velocity of 50m/s so it will cover 400m in 8 seconds ( 400/50 =8 ). Now you can use this to find the distance covered by the balls as mentioned in the previous answers.
```
11 days ago
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### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions