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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.

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.

Grade:12th pass

3 Answers

sumit hajare
28 Points
4 years ago
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 
Vikas TU
14149 Points
4 years ago
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
Nitin Kumar
26 Points
4 years ago
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.
 

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