Question

a stone is allowed to fall from the top of a tower 100m high and at the time another stone is projected vertically upwards from the ground with the velocity of 25m/s. Calculate when and where the two stones will meet​

Answer 1

Solution :-

• Height of the tower(h) = 100m

Stone 1 is allowed to fall from the top hence '

• initial velocity (u) = 0 m /s

• Distance covered by the stone 1 = y

As the stone moves uniformly under gravity we can apply the second equation of motion ,

➽ s = ut + gt² /2

➽ y = 10x t² /2

➽ y = 5t² ......(1)

Stone 2 is projected vertically from the ground hence ,

• initial velocity = 25 m/s

• Distance traveled by the stone = 100 -y

As the stone moves uniformly under gravity we can apply the second equation of motion ,

➽ s = ut + gt² /2

➽ 100 -y = 25t - 10x t² /2

➽ 100 -y = 25t - 5t² .....(2)

On adding equation (1) and (2) we get ,

100 -y +y = 25t - 5t² + 5t²

100 = 25 t

t = 100 / 25

t = 4 s

Now let's substitute the value of t in eq (1) ,

➽ y = 5t ²

➽ y = 5 x 16

➽ y = 80 m

Distance of the stone from the ground

➽ y' = 100 - y

➽ y' = 100 - 80

➽ y' = 20 m

The two stones will meet at a distance of 20m from the ground at t = 4 s

Answer 2

Answer:

ANSWER

Let the stones meet at point A after time t.

For upper stone :

u

′

=0

x=0+

2

1

gt

2

x=

2

1

×10×t

2

⟹x=5t

2

............(1)

For lower stone :

u=25 m/s

100−x=ut−

2

1

gt

2

100−x=(25)t−

2

1

×10×t

2

⟹100−x=25t−5t

2

............(2)

Adding (1) and (2), we get

25t=100

⟹t=4 s

From (1),

x=5×4

2

⟹x=80 m

Hence the stone meet at a height of 20 m above the ground after 4 seconds.

solution