Question

A ball falls freely from rest the ratio of the distance travelled in first , second , third and fourth sec is

Answer 1

$\huge{\mathfrak{Question:-}}$

A ball falls freely from rest the ratio of the distance travelled in first , second , third and fourth sec is

$\huge{\mathfrak{Solution:-}}$

Distance travel is time t is:

$\bold{s = ut + 1/2at^{2}}$

Initial velocity u = 0

$\bold{s = 1/2at^{2}}$

Distance travel in 1 second = $\bold{s_{1}= 1/2\times 9.8\times 1^{2}=4.9\;m}$

similarly distance travelled in 2, 3, 4 seconds are:

$\bold{s_{2}= 19.6\;m} \\ \\ \\ \bold{s_{3} = 44.1\;m} \\ \\ \\ \bold{s_{4} = 78.4\;m}$

Now distance travelled in first (D1), second(D2), third(D3) and fourth(D4) second are:

$\bold{D_{1}= s_{1}=4.9\;m}$

$\bold{D_{2}=s_{2}-s_{1}=19.6-4.9 =14.7\;m}$

$\bold{D_{3}=s_{3}-s_{2}=44.1-19.6 = 24.5\;m}$

$\bold{D_{4}=s_{4}-s_{3}=78.4 = 44.1= 34.3\;m}$

$\boxed{\sf{Ratio\;\;D_{1}:D_{2}:D_{3}:D_{4} = 1:3:5:7}}$

Answer 2

body is falling from rest so its initial velocity = u=0

accleration due to gravity = g

distance covered in first second = ut + at^2/2                   (u=0)

                                          s1  =gt^2/2=g/2     .......1                   (t=1 , a=g)

distance covered from t=0 to t=2 sec is

                              =ut+at^2/2=at^2/2                         (u=0)

                                            =2g                              (t=2)

distance covered in second second = s2 =  2g - g/2=3g/2  ..................2

divide 1 & 2

s1/s2 = 1/3 ans