Question

. 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.​

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:

Expert Verified

(100-S) = distance travelled by the projected stone. Therefore, Two stones will meet after 4 s. = 80 m.