Question

a stone is droppedfrom height of 100m fromgroundand at the sametime another stone isthrownup withvelocity of 25m/s. calculate when and where both stones meet? also had the distancetravelled by bothstones then meet

Answer 1

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 stone dropped down be s1. So the stone thrown up is s2.

s1 => d = 100-x

          g = 9.8 m/s²

          u = 0

We know, s= ut + 1/2at²

Putting values,

100-x = 4.9t²                  ....(1)

 

s2 => d = x

          g = -9.8 m/s²

          u = 25 m/s 

We know, s= ut + 1/2at²

Putting values,

x = 25t -4.9t²                  ....(2)

Adding (1) and (2),

100-x +x = 4.9t² + 25t - 4.9t²

t = 4 secs.

Puuting t = 4 in (1),

100-x = 4.9t²

100-x =19.6

 

x = 80.4 m

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