Question

a stone dropped from the top of a long pole crosses two points separated by 20m in 1s. find the distance of first point from the top of the pole. (g=10ms)

Answer 1

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.

Answer 2

The correct answer is 11.25 m.

Given: Distance between two points = 20m.

Time taken to cross the two points = 1 sec.

To Find: Distance of first point from the top of the pole.

Solution:

Let $v_{1}$ be the velocity at the first point and $v_{2}$ be the velocity at the second point while crossing the points on the pole.

As we know that v = u + at

v₂ = $v_{1}$ + g t

v₂ = v₁ + 10 × 1

v₂ = v₁ + 10 m/s  (equation 1)

Also, we know that $v^{2}- u^{2}$ = 2 as.

$v_{2} ^{2} -v_{1} ^{2}$ = 2 g s

$(v_{1}+10) ^{2} -v_{1} ^{2}$ = 2 × 10 × 20

$v_{1} ^{2}+20v_{1}+100 -v_{1} ^{2}$ = 400

20v₁ = 400 - 100 = 300

v₁ = $\frac{300}{20}$ = 15 m/s

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.

v₁ = u + g t

15 = 0 + 10 t

t = $\frac{15}{10}$ = 1.5 sec

Distance = s = $ut+\frac{1}{2}at^{2}$

s = $0+\frac{1}{2}*10*1.5^{2}$ = 5× 2.25 = 11.25 m

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

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