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Grade: 12th pass
        
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.
3 months ago

Answers : (3)

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 
3 months ago
Vikas TU
9499 Points
							
Dear student 
A ball dropped from the roof of height 400m Let after t time it meets with another ball
So,
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 m
So, height of collision from ground or surface of tower is (400-320)=80 m
3 months 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.
 
2 months ago
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