Question

In the last second of free fall a body covered 3/4 of its total path. Then the height from which the body is released is

Answer 1

Let the total time in free fall be t. Then, the distance covered just before the last second would be 1/4th the total distance, or t-1 seconds.

h(t)=1/2 gt²

h(t-1)=1/2g(t-1)²

1/4(1/2 gt²)=1/2 g(t-1)²

g/2t²=2g(t²-2t+1)

t²=4t²-8t+4

3t²-8t+4=0

(3t-2)(t-2)=0

t=2/3 or 2

We can ignore the 2/3rds seconds, and get a value for t as 2 seconds. So, the body fell for 2 seconds, and covered 4.9(4), or 19.6m

Answer 2

Answer:

19.6 m

Explanation:

We have to take the last second of the fall to be t−1 where t is the total time of fall.

Consider h is the total height. For a freely falling body, we take initial velocity to be zero. So, u=0.Also consider t is the time of fall and the acceleration to be g=10ms−2. Now, applying the equations of motion:

s=ut+12at2

h=0×t+12(10)t2⇒h=5t2 …(1)

Now, in the last second the initial velocity u becomes u+g(t−1) and time t becomes t−1 and we are provided with that the total height in the last second is 34th of total height. So, height hchanges to 34h

Now, the equation of motion be:

34h=[u+g(t−1)](t−1)+12g(t−1)2

Using (1) and solving

34(5t2)=32(10)(t−1)2

⇒t2=4(t−1)2⇒3t2−8t+4=0⇒3t2−6t−2t+4=0⇒3t(t−2)−2(t−2)=0⇒(t−2)(3t−2)=0

Hence, we got t=2,23

But time cannot be in fraction. Hence, time of fall is 2sec

Now putting the value in (1)

h=5(2)2⇒h=20

Total height will be 20m when we take the g=10ms−2.

And the height by taking g=9.8ms−2.

∴h=12×9.8×22=19.6

Hence, the total height is 19.6m