Physics, asked by sivaroswald, 1 year ago

A freely falling body describes a distance x in first 2 seconds and a diatance y in next 2 seconds .Then show that y=3x

Answers

Answered by MayankTamakuwala1296
38
Object is falling freely

In first two seconds it will cover distance of

d = ut + [(at²)/2]

Here, d = x , u = 0, t = 2s, a = g = 9.8 m/s²

x = (0)(2) + [(9.8 * 2²)/2]

x = 0 + (9.8 * 4)/2

x = 19.6 m ---------------------(1)

So in first two seconds it will cover 19.6 m distance

For first 2 seconds that is till it covers x meter distance it's velocity at that point will be

v = u + at

Here, a = g = 9.8 m/s² , t = 2s , u = 0 m/s

v = 0 + (9.8 * 2)

v = 19.6 m/s

So, while it start for next two seconds its initial velocity will be 19.6 m/s

Now, we have to find y that is distance covers in next 2 seconds will be

d = ut + [(at²)/2]

Here, d = y , u = 19.6 m/s, t = 2s, a = g = 9.8 m/s²

y = (19.6)(2) + [(9.8 * 2²)/2]

y = 39.2 + (9.8 * 4)/2

y = 58.8 m -----------------(2)

So it will cover y = 58.8 m distance in next 2 seconds.

Now, Dividing equation 2 with equation 1

 \frac{y}{x}  =  \frac{58.8}{19.6}

 \frac{y}{x}  = 3

y = 3x

Hence proved
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Answered by drkruthikanaidu1599
17

Initial velocity u=0(Bz it is at rest)

Distance travelled in first two seconds is

S = ut +1/2at^2

X = (0*2) + 1/2a(2)^2

X = 0 + 1/2a(4)

X = 0+2a

X = 2a ----->1st equation

Distance travelled in first 4 seconds is

S = ut + 1/2at^2

X + Y = (0*4) + 1/2a(4)^2

X+ Y = 0 + 1/2a(16)

X + Y = 0 + 8a

X + Y = 4(2a) (bz frm equation i.e x =2a)

X + Y = 4X

Y =4X - X

Y =3X

Hence proved

Sorry as I couldn't write square of t so I used t^2

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