Question

A stone is dropped from the top of the tower 100 m height. At the same instant another stone is thrown vertically upward from the base of the tower with the velocity of
25 m/s . when and where will two stones meet? (Given: g= 10 m/s²)​

Answer 1

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.

Answer 2

Answer:

Let the ball dropped down is a and the ball thrown up is b. also the stone thrown upwards cover a distance of x and the other covers a distance of (100 –x).

For ball a

u=0

g=10m/s

2

d=(100−x)

Using the equation

s=ut+

2

1

at

2

100−x=5t

2

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

For ball b

d=x

g=−10m/s

2

u=25m/s

s=ut+

2

1

at

2

x=25t−5t

2

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

Solving equation (1) and (2)

100=25t

t=4seconds

Put the value of t in equation (1)

x=100−80

x=20m

They will meet at distance of 80 m from the ground after t = 4 seconds

Explanation:

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