Question

a stone dropped from some height crosses the top and bottom ends of a vertical pole of length 20m in 1sec , the distance of top end of pole from point of release
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Answer 1

Answer:

Explanation:

Hey mate here is your answer:

s = 20 m ; a = g = 10 m/s^2 ; t = 1 s

Let v1 be the velocity at the first point and v2 be the velocity at the second point while crossing the points on the pole.

As we know that v = u + at

so, v2 = v1 + g t

or, v2 = v1 + 10 × 1

or, v2 = v1 + 10 m/s___________(1)

Also, we know that v^2 - u^2 = 2 a s

so, v2^2 - v1^2 = 2 g s

or, (v1 + 10)^2 - v1^2 = 2 × 10 × 20

or, v1^2 + 20v1 + 100 - v1^2 = 400

or, 20v1 = 400 - 100 = 300

so, v1 = 300/20 = 15 m/s

so, v2 = v1 + 10 = 15 + 10 = 25 m/s

so, as velocity at the top of the pole when it was dropped will be u = 0 m/s

so, v1 = u + g t

or, 15 = 0 + 10 t

so, t = 15/10 = 1.5 sec

so, distance, s = u t + 1/2 g t^2

or, s = 0 + 1/2 × 10 × (1.5)^2 = 5× 2.25 = 11.25 m

so, 11.25 m is the height from the first point to the top.

Hope it helps you.

Answer 2

Answer:

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