Question

a stone is dropped from the top of a tall cliff and 1 sec later a second stone is throw vertically downward with a velocity of 20 m per sec . how far below the top of the cliff will the second stone overtake the first

Answer 1

Answer:

Let after t seconds second stone overtake the first stone at distance h from the cliff.

Using s=ut+  

2

1

​  

gt  

2

,  

For the first stone,

⇒−h=−  

2

1

​  

g(n+t)  

2

  ⇒h=  

2

1

​  

g(n+t)  

2

  ...1

For the second stone,

⇒−h=−ut−  

2

1

​  

gt  

2

  ⇒h=ut+  

2

1

​  

gt  

2

   .....2

Equating equations 1 and , ⇒ut+  

2

1

​  

gt  

2

=d  

2

1

​  

g(n+t)  

2

  ⇒t=  

2(u−gn)

gn  

2

 

​  

 

Putting t in equation 1, ⇒h=(  

2

g

​  

)[n+  

2(u−gn)

gn  

2

 

​  

]  

2

  ⇒h=(  

2

g

​  

)[n  

gn−u

(  

2

gn

​  

−u)

​  

]  

2

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