Mon May 23, 2011 By: Hiba Vallangara

From the top of the tower 100m in height a ball is dropped and at the same time another ball is projected vertically upward from the ground with velocity of 25m/s. Find when and where the 2 balls meet(a=9.8m/s2

Expert Reply
Mon May 23, 2011

They will meet instinctively. Here's how.

Lets take distance covered by the stone thrown upwards to meet the other as x. Then the other stone covers a distance of (100-x)m.

Let the ball dropped down be b1. So the ball thrown up is b2.

b1 => d = 100-x

          g = 9.8 m/s sq.

          u = 0

We know, s= ut + 1/2at sq.

Putting values,

100-x = 4.9t sq.                  ....(1)


b2 => d = x

          g = -9.8 m/s sq.

          u = 25 m/s 

We know, s= ut + 1/2at sq.

Putting values,

x = 25t -4.9t sq.                  ....(2)

Adding (1) and (2),

100-x +x = 4.9t sq. + 25t - 4.9t sq.

t = 4 secs.

Puuting t = 4 in (1),

100-x = 4.9t sq.

100-x =19.6


x = 80.4 m

So, they meet at  80.4 m from ground after 4 seconds. 

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