. A stone is allowed to fall from the top of a tower 100 m high and at the same time another stone


is projected vertically upwards from the ground with a velocity of 25 m/s. Calculate when and where the two stones will meet.

 

Best Answer

Sol. ||| Firstly, for the stone falling from top of tower:

Height, h = (100 - x)

Initial velocity, u = 0

Time, t =?

g = 9.8 m/s2

We know that,

a stone is allowed to fall

Now, for stone projected vertically upwards:

Height, h = x

Initial velocity, u = 25 m/s

Time, t =?

g = -9.8 m/s2 (As the stone goes up)

We know that,

top of tower

Therefore, the 2 stones will meet after a time of 4 seconds.

Now, putting the value of t in (i), we get:

 100 – x = 4.9 × (4)2

 100 – x = 4.9 × 16

 100 – x = 78.4

 100 – 78.4 = x

 x = 21.6 m

Therefore, the two stones will meet at a height of 21.6 m above the ground.

 

Talk to Our counsellor