Question

Two stones one is dropped freely from the top of a building of loom and the next is thrown vertically upward with 20m/s at the same instant. Now find out where and when meet each other?​

Answer 1

Let's assume that the acceleration due to gravity is `g = 9.8 m/s²` and that the building is tall enough that the stones do not reach the ground during the time period we are considering. Let `h1(t)` be the height of the first stone (the one dropped freely) at time `t`, and let `h2(t)` be the height of the second stone (the one thrown upward) at time `t`. Since the first stone is dropped freely, its height at time `t` is given by the equation `h1(t) = 100 - (1/2)gt²`. Since the second stone is thrown upward with an initial velocity of `20 m/s`, its height at time `t` is given by the equation `h2(t) = 100 + 20t - (1/2)gt²`.

The stones will meet each other when their heights are equal, i.e., when `h1(t) = h2(t)`. Substituting the expressions for `h1(t)` and `h2(t)` into this equation, we get:

`100 - (1/2)gt² = 100 + 20t - (1/2)gt²`

Solving this equation for `t`, we find that `t = 5 s`. Substituting this value of `t` into either of the equations for `h1(t)` or `h2(t)`, we find that the height at which the stones meet is `h1(5) = h2(5) = 100 + 20(5) - (1/2)(9.8)(5)² ≈ 50 m`.

Therefore, **the two stones will meet each other at a height of 50 meters above the ground, 5 seconds after they are released**.

Answer 2

Step-by-step explanation:

As per the question, both the stones will meet after

3

sec

.

Let, total height of the building be

H

and the distance covered by dropped and thrown stone be

H

1

and

H

2

respectively, we have

H

=

H

1

+

H

2

So, from equation of motion we have

s

=

u

t

+

1

2

a

t

2

⇒

H

1

=

0

(

3

)

+

1

2

×

10

×

(

3

)

2

⇒

H

1

=

45

m

also,

H

2

=

(

20

)

(

3

)

−

1

2

×

10

×

(

3

)

2

⇒

H

2

=

15

m

Hence, total height of the building is

H

=

45

+

15

=

60

m