Question

147. A stone is dropped from the top of a tower towards
the earth. Another stone is released from the same
point exactly one second later. What is the separation
between the two stones, two seconds after the release
of the second stone is (take g = 10 m/s)

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Answer 1

SOLUTION.

As we can see that the time for the motion of the first stone is more hence it is clear that the second stone can not overtake first stone.

Total time for motion of first stone is 3 sec and for second stone it is 2sec

• For first stone

Let this stone cover h height after the 3sec.

Initial velocity = 0

Height = h

Time = 3 sec

Than acc to eq of motion.

h = ut + (1/2)gt²

h = 0 + 5(3)²

h = 45m

• For second stone

let the second stone covers H height after its release means after 2sec

Initial velocity = 0

height = H

time = 2sec

than acc to eq of motion.

H = ut + (1/2)gt²

H = 0 + 5(2)²

H = 20m

Separation = h - H

SEPARATION = 25m

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