Question

A stone is dropped from the top of a cliff of height h, n second later, a second stone isprojected downwards form the same cliff with a vertically downward velocity u. Show thatthe two stones will reach the bottom of the cliff together, if 8h (u gn) 2 = gn 2 (2u gn)

Answer 1

Explanation:

We are given the values of time, but the values of the Sol. The area under velocity-time graph represents distance

and displacement of the particle,

position are to be obtained from the graph corresponding to the

1. Distance travelled by the particle is the total length of the

given time interval under consideration, We know that slope of path travelled by the particle, Hence distance travelled by the

particle

the displacement-time graph represents velocity,

S = Total sum of arca under the graph

X2 -Xl

= 1 2 + 4) x 4 + 1 + 4) x 2= 12 + 6 = 18 m

Formula to be used: Vav = - - - where Xl and X2 are the -( -(2

22

t2 - t]

2. The displacement of the particle is the change of position,

initial and final positions" respectively and I, and I, are initial The displacement of the particle is positive for area above

and final values of time, respectively, the time axis and negative for the area below the time axis..

Hence displacement of the particlc is

= x,1. Here X, We are given the values of time, but the values of the Sol. The area under velocity-time graph represents distance

and displacement of the particle,

position are to be obtained from the graph corresponding to the

1. Distance travelled by the particle is the total length of the

given time interval under consideration, We know that slope of path travelled by the particle, Hence distance travelled by the

particle

the displacement-time graph represents velocity,

S = Total sum of arca under the graph

X2 -Xl

= 1 2 + 4) x 4 + 1 + 4) x 2= 12 + 6 = 18 m

2. The displacement of the particle is the change of position,

initial and final positions" respectively and I, and I, are initial The displacement of the particle is positive for area above

and final values of time, respectively,

Therefore the required average velocity is given by

10 - 0 -l

Vav = 2 _ 0, = 5 ms

2. Here XI = 10m, X2 =5 m, I, = 2 S, 12 = 4:s L'>x = I + 4) x 4 - 1 + 4) x 2 = 12 - .

So the required value of average velocity is given by 6= 6m