Falling Body ! - Physics

x = 1/2 g t^2

1/2 x = 1/2 g t^2 - 1/2 g (t-1)^2

Solve for t.

Or since the question is multiple choice, make an estimate:

In 1 second, an object falls 4.9 m. For simplicity let's call that distance L. In 1 second, an object falls distance L.

In 2 seconds, an object falls distance 4L. In the first second it fell L, so in the second second it fell 3L.

In 3 seconds, an object falls distance 9L. In the first two seconds it fell 4L, so in the last second it fell 5L.

In 4 seconds, an object falls distance 16L. In the last second it fell 7L.

The last second being half of the total occurs somewhere between 3 and 4 seconds for the total time falling.

Let s be total path distance & t be the total duration of the fall of the body in seconds.

S(t)=s=ut+(1/2)gt^2=(1/2)gt^2.................(I) (since u=0)

S(t-1)=(1/2)g(t-1)^2.......................(II)

(I)-(II) gives,

S(t)-S(t-1)=(1/2)g(2t-1)=s/2 (Distance travelled in the last second of its free fall)

i.e. s=g(2t-1)

substituting value of s in (I), we get

g(2t-1)=(1/2)gt^2 which implies,

t^2-4t+2=0 which is quadratic in t.

Hence, t=(4+√(4^2-4×2×1))/2=(4+√8)/2=(4+2√2)/2=(2+√2)=3.414≈3.4

Since t>1, t=(2-√2) is not considered.

Therefore, t≈3.4

I even didn't understand ur question. what this last sec of free fall?

The distance travelled by the body in the $n^{th}$ second is given by $S_n=u+\dfrac{g}{2}(2n-1)$ . We also know that $S=\dfrac{1}{2}gt^2$ It is given that at the last second it covers $\dfrac{S}{2}$ $\dfrac{gt^2}{4}=\dfrac{g}{2}(2t-1) \implies t^2-4t+2=0 \implies t=2+\sqrt{2} \approx 3.414 \ \text{sec}$

in the first sec the body travels @ half the speed , it would travel in the next second, so approx double the last sec ie 2 secs and the last sec ie about 3

We know ,

Time of Fall 'T' = $\sqrt{2H/g}$ - $\boxed{1}$

Displacement of a body in the nth second of its motion :

'S' = $u + a\times(n -\frac{1}{2})$

We know in the last second :

T = $\frac{H}{2}$

Displacement of the body in the last second :

$\frac{H}{2}$ = $g\times(T - \frac{1}{2})$

Solving the equation we get :

T = $\frac{H}{2g} - \frac{1}{2}$ - $\boxed{2}$

Substituting the value of 'T' in equation - $\boxed{1}$

We get $H^{2}/4g^{2} - \frac{3H}{4g} + \frac{1}{4}$ = 0

Solving the Quadratic equation we get H = 3.23 u

T = $\sqrt{12.92}$ = 3.6 approx

Note : Here I used g = 10 and by using g = 9.8 we could get a value very close to 3.4s

if u solve this prblm logically the body travels half of total its path in last second of its free falll it means that aceeleration will be more due to gravity timee will be less
the more speed lesser the time

USE EQUATION OF MOTION