Question

A stone is dropped from a cliff at time t = 0 and strikes the ground below.

Answer 1

Answer:

S=

2

1

gt

2

........(1)

And that of the other stone is

S=u(t−n)+

2

1

[g(t−n)

2

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

Since both the stones meet at the distance so equation (1)and equation(2) will be equal

2

1

gt

2

=u(t−n)+

2

1

[g(t−n)

2

]

gt

2

=2ut−2un+gt

2

+gn

2

−2gnt

t(2gn−2u)=gn

2

−2un

t=

(gn−u)

n(

2

gn

−u)

now putting value of t in equation(1)

s=

2

g

⎣

⎢

⎡

gn−u

2

[n(

2

gn

−u)]

⎦

⎥

⎤

2

S=

2

g

⎣

⎢

⎡

(gn−u)

2

n(

2

gn

−u)

⎦

⎥

⎤

Explanation:

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