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  • A ball is allowed to fall from the top of a tower of 200 meter high. At the same instant another ball is thrown vertically upwards from the bottom of the tower with a velocity of 40 meters per second. When and where the two balls meet

A ball is allowed to fall from the top of a tower of 200 meter high. At the same instant another ball is thrown vertically upwards from the bottom of the tower with a velocity of 40 meters per second. When and where the two balls meet

Answers (1)

Let the balls meet at a height h from the ground. The dropped ball travels a height 200-h and the ball thrown vertically upward travels height h. Let t be the time at which they meet. 

The height travelled by dropped ball

200-h=0.5\times g\times t^2.....(1)

The height travelled by ball thrown vertically upward

h=40t-0.5\times g\times t^2.....(2)

Add (1) and (2)

200=40t

t = 5 sec

Substitute the value of t in equation (1) then 

h=200-0.5\times10\times5^2=75m

Posted by

Safeer PP

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