Question

19. A stone is thrown vertically downwards from the top of a cliff 175 metre high with a velocity of 20 m/s.
Another stone is projected vertically upwards from the base of the cliff with a velocity of 50 m/s. When
and where will the two stones meet?

Ans:
(After 2.5 second at a height of 94.375 m above the
ground)

plz give correct explanation of this Qn . You can cross check it with the ans I gave .
I want the explanation of the Ans given above​

Answer 1

Explanation:

Let, they meet after time 't' a 'x' distance from ground.

* for 1st stone ( thrown vertically downward)

U= 20 m/s

S=175-x

a=g

t= t

* s= Ut + 1/2gt×t

=>175-x= 20 t+ 4.9 t×t -------(i)

*for 2nd stone

u= 50 m/s

S= x

a= g

t=t

* S = ut - 1/2 gt×t

=>x= 50t - 4.9 t× t--------(ii)

so,

175-x= 20t + 4.9t× t--------(i)

x= 50t-4.9 t×t --------(ii)

Adding,

175=70t

=> t=175/70

=>t= 2.5 sec

putting the value of 't' in (ii),

x=50(2.5)-4.9(2.5×2.5)

=125-30.625

=94.375 m.

so,

the stones meet each other ater 2.5 sec at 94.375m

above the ground. (Answer)

*

Answer 2

Given:

• Height of the cliff = 175 m

• initial velocity of stone thrown downward from top of cliff , u₁ = 20 ms⁻¹

• initial velocity of stone thrown upward from base of cliff , u₂ = 50 ms⁻¹

To find:

• when and where stones will meet

Formula required:

• second equation of motion

        s = u t + 1/2 a t²

(where s is the distance travelled by body , u is its initial velocity , a is its acceleration and t is time )

Solution:

Let,

Two stones meet  after a time t

also, Let distance travelled by stone thrown downward before meeting be s₁

and Distance travelled by stone thrown upward before meeting be s₂

Then,

taking value of gravitation due to gravity 9.8 ms⁻²

Using second equation of motion for stone thrown downward

→ s₁ = (20) t + 1/2 (9.8) t²

→ s₁ = 20 t + 4.9 t²  

Using second equation of motion for stone thrown upward

→  s₂ = (50) t + 1/2 (-9.8) t²

(taking value of acceleration due to gravity negative because stone is thrown upward)

→  s₂ = 50 t - 4.9 t²  

Now,

Since height of cliff is 175 m

therefore, total distance travelled by both stones will be equal to 175 m

so,

→ s₁ + s₂ = 175

→ 20 t + 4.9 t² + 50 t - 4.9 t² = 175

→ 70 t = 175

→ t = 2.5 s

Now,

putting the value of t = 2.5 s in the equation s₂ = 50 t - 4.9 t² (because s₂ gives the distance from ground where they will meet)

→ s₂ = 50 t - 4.9 t²

→ s₂ = 50 (2.5) - 4.9 (2.5)²

→ s₂ = 125 - 30.625

→ s₂ = 94.375 m

Therefore,

The two stones will meet after 2.5 seconds at a height of 94.375 metres from the ground.