Question

If a stone when released from rest, travels half of its total
path in the last second of its fall. Height from which the body
falls is nearly

Answer 1

Explanation:

Correct Answer is: (2) 3.41 s, 57 m

0.59 s is physically unacceptable as it gives the total time t taken by the body to reach ground is lesser than one sec while according to the given problem time of motion must be greater than 1 s.

Answer 2

Given:

Stone is released from rest and travels half of its total path in the last second of its fall.

To find:

Height from which the body falls.

Solution:

We assume that the total time = t sec

Distance travelled in t sec:

S = ut + 1/2at²

As initial velocity u =0

S = 1/2gt²

The distance fallen in (t-1) sec will be:

S' = 1/2g(t-1)²

Distance fallen in last second will be:

S- S' = 1/2g [t² - (t-1)²] =1/2g(2t - 1)

As given in the question the stone travels half of the distance in the last second:

S - S' = S/2

S/2 = 1/2 g (2t - 1)

Putting the value of S we get:

1/4 gt² = 1/2 g(2t - 1)

1/2t² = 2t-1

t² = 4t -2

t² - 4t +2 = 0

t = [4±√4²-4×2]/2

t = 2±√2

Therefore t = 0.59 sec or 3.41 sec

The time must be greater than 1 sec as the question says that the stone travels half of the distance in the last second.

∴ t = 3.41 sec

Since h = 1/2gt² = 1/2*9.8*(3.41)² = 57 m

Therefore the height is 57 m.