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Grade 12IIT JEE Entrance Exam

From the top of the tower of height 400 m, a ball is dropped by a man. Simultaneously from the base of the tower, another ball is thrown up with velocity 50 m/s. At what distance will they meet from the base of the tower ?

Kiran Alex Johnson

11 Years agoGrade 12

1 Answer

Sumit Majumdar

11 Years ago

Dear student,
Let us assume that the balls meet at a distance of x m from the base of the tower.
Hence the velocity achieved by the ball by the time it reaches the distance (400-x) from the top is:
$v^2=u^2+2gh\Rightarrow v^2=0^2+2\times9.8\(400-x\)\Rightarrow v=\sqrt 19.6\(400-x\)$ Lets assume that it takes t sec to cover this distance, hence, we have:
$400-x=0\times t+\frac{1}{2}gt^2$
Similarly for the ball thrown from the bottom, we have:
$x=50t-\frac{1}{2}gt^2$
so, we get:
$400=50t\Rightarrow t=8 sec$
hence,
$x=50\times 8-\frac{1}{2}\times 9.8\times 8^2=400-313.6=86.4 m$
Regards

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