Question

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A stone is dropped from rest from the top of a cliff . a second stone is thrown vertically down with a velocity of 30m/s two seconds later. at what distance from the top of a cliff do they meet ?

1 :) 60 m

2:) 120 m

3:) 80 m

4:) 44m

Answer 1

answer :-


the two stones meet at distance S from the top of cliff t. seconds after first stone is dropped.


for 1st stone

s = 1 / 2 gt^ 2 ;

for second stone

s = u ( t - 2 ) + 1 / 2 g (t -2) ^2


since,

1 / 2 gt^2 = ut - 2u + 1 /2 gt^2 - 2gt +2g


0 = (u -2g) t -2 (u - g) ;


t = 2 (u - g ) / u - 2g

=> 2 ( 30 - 10) / 30 - 20

= 4 s


distance S at which they meet = 1 / 2 gt^2 = 1 /2 × 10 × 16 :-

= 80 m from top of the cliff.


option ( c).

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Answer 2

Answer-:

Distance S which they meet= 1/2

gt^2= 1/2×10×16

=>80m

=> (3) is correct

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