Question

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

Answer 1

Solution :

The to a stone's meet at a distance S from top of Cliff t
seconds after first stone is dropped.


For 1st stone S = 1/2 gt²

For 2nd stone S = u( t -2 ) + 1/2 g(t-2)²


now equate the both distance

1/2 gt² = ut – 2u + 1/2 gt² – 2gt + 2g

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

t = 2( u - g ) /u - 2g = 2(30 - 10 ) /30 - 20 = 4 sec

t = 4 second

distances S at which they meet = 1/2 * gt²

= 1/2 * 10 * 16

= 80 m from top of Cliff

_________________________
❤BE BRAINLY ❤

Answer 2

Answer:

Both the stones will meet at a distance of 80m from the top of the cliff.

Explanation:

After the first stone is dropped, let the second stone be dropped at a distance of S from the cliff's peak t.

So, for 1st stone S = 1/2 gt²

2nd stone will be S = u(t -2) + $\frac{1}{2}$ g(t - 2)²

now let's equate the both distance 1/2 gt² = ut – 2u + $\frac{1}{2}$ gt² – 2gt + 2g

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

t = 2( u - g )/u - 2g = 2(30 - 10)/30 - 20

= 4 sec

So, now t = 4 second

Therefore, distances S at which they will meet = $\frac{1}{2}$ * gt²

= $\frac{1}{2}$ * 10 * 16

= 80m from the top of the Cliff

Thus, both the stones will meet at 80m from the top of the cliff.