Question

a stone falls from rest.the distance covered by the stone in last second of its motion equals the distance covered by it during the first three seconds of its motion.how long does the stone take to reach the ground​

Answer 1

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

• Distance travelled in the Last second = Distance Travelled in First 3 seconds.
• Initial velocity (u) = 0 m/s.

$\huge{\bold{\underline{Explanation:-}}}$

$\rule{300}{1.5}$

From, Second kinematic equation.

$\large{\boxed{\tt S = ut + \dfrac{1}{2}at^2}}$

Substituting the values,

$\large{\tt \leadsto S = 0 \times t + \dfrac{1}{2} \times 10 \times (3)^2}$

• t = 3 seconds.
• a = 10 m/s².

$\large{\tt \leadsto S = 0 + \dfrac{1}{2} \times 10 \times 9}$

$\large{\tt \leadsto S = \dfrac{1}{2} \times 10 \times 9}$

$\large{\tt \leadsto S = \dfrac{1}{\cancel{2}} \times \cancel{10} \times 9}$

$\large{\tt \leadsto S = 5 \times 9}$

$\large{\boxed{\tt S = 45 \: meters}}$

So, The Distance travelled in First 3 seconds and Last second is 45 meters.

$\rule{300}{1.5}$

$\rule{300}{1.5}$

From Distance in nth Second formula,

$\large{\boxed{\tt S_n = u + \dfrac{a}{2}(2n - 1)}}$

∵$\large{\tt S_n = S}$

Substituting the values,

$\large{\tt \leadsto 45 = 0 + \dfrac{10}{2}(2n - 1)}$

$\large{\tt \leadsto 45 = 0 + \cancel{\dfrac{10}{2}}(2n - 1)}$

$\large{\tt \leadsto 45 = 0 + 5(2n - 1)}$

$\large{\tt \leadsto 45 = 5(2n - 1)}$

$\large{\tt \leadsto \dfrac{45}{5} = (2n - 1)}$

$\large{\tt \leadsto \cancel{\dfrac{45}{5}} = (2n - 1)}$

$\large{\tt \leadsto 2n - 1 = 9}$

$\large{\tt \leadsto 2n = 9 + 1}$

$\large{\tt \leadsto 2n = 10}$

$\large{\tt \leadsto n = \dfrac{10}{2}}$

$\large{\tt \leadsto n = \cancel{\dfrac{10}{2}}}$

$\huge{\boxed{\boxed{\tt n = 5 \: seconds}}}$

So, the Time taken to reach the ground is 5 seconds.

$\rule{300}{1.5}$

Answer 2

$\Huge{\underline{\underline{\blue{\mathfrak{Answer :}}}}}$

We are Given

• Distantce travelled in last second and third second are equal.
• And stone fell from rest, means initial velocity (u) = 0 m/s
• And Gravitational force be 10 m/s²

$\rule{100}{2}$

We have formula :-

$\LARGE \implies {\underline{\boxed{\green{\tt{S \: = \: ut \: + \: \frac{1}{2} \: at^2}}}}}$

Putting Values

$\Large \leadsto {\sf{S \: = \:=(0)(3) \: + \: \frac{1}{2} \: (10)(3)^2}}$

$\Large \leadsto {\sf{S \: = \: 0 \: + \: \frac{1}{2} \: (10)(9)}}$

$\Large \leadsto {\sf{S \: = \: 0 \: + \: \frac{1}{\cancel{2}} \: \cancel{90}}}$

$\Large \implies {\boxed{\red{\sf{S \: = \: 45}}}}$

$\rule{200}{2}$

We have formula for nth second,

$\LARGE \implies {\boxed{\boxed{\green{\tt{S_{n} \: = \: u \: + \: \frac{a}{2} \: (2n \: - \: 1)}}}}}$

Putting Values

$\Large \leadsto {\sf{45 \: = \: 0 \: + \: \frac{\cancel{10}}{\cancel{2}} \: (2n \: - \: 1)}}$

$\Large \leadsto {\sf{45 \: = \: 5 (2n \: - \: 1)}}$

$\Large \leadsto {\sf{45 \: = \: 10n \: - \: 5}}$

$\Large \leadsto {\sf{45 \: + \: 5 \: = \: 10n}}$

$\Large \leadsto {\sf{50 \: = \: 10n}}$

$\Large \leadsto {\sf{n \: = \: \frac{\cancel{50}}{\cancel{10}}}}$

$\Large \implies {\boxed{\red{\sf{n \: = \: 5}}}}$