Question

a stone is dropped from a hundred metre cliff. What is the velocity of the stone after 3 seconds ? When does the stone hit the ground? With what velocity does the stone hit the ground?​

Answer 1

Answer:

Given:

Height = 100 metres

Initial Velocity = 0 m/s

To find:

• Velocity of stone after 3 seconds
• Velocity with which stone hits the ground
• Time after which the stone hits ground

Calculation:

Velocity after 3 seconds:

$\therefore \: v = u + gt \\ = > v = 0 \: + 10 \times 3 \\ = > v = 30 \: metres \:{s}^{-1}$

Velocity on reaching ground:

$\therefore \: {v}^{2} = {u}^{2} + 2gh \\ = > {v}^{2} = {0}^{2} + 2 \times 10 \times 100 \\ = > {v}^{2} = 2000 \\ = > v = \sqrt{2000 } \\ = > v = 44.72 \: m {s}^{ - 1}$

Time after which body reaches ground :

$\therefore \: h = ut + \frac{1}{2} a {t}^{2} \\ = > 100 = 0 + \frac{1}{2} \times 10 \times {t}^{2} \\ = > {t}^{2} = 20 \\ = > t = \sqrt{20} \\ = > t = 4.47 \: sec.$

Answer 2

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

Given :

• Height = 100 m
• Intial Velocity (u) = 0 m/s

Solution :

We have formula

$\large \bigstar {\boxed{\sf{v \: = \: u \: + \: at}}} \\ \\ \\ \implies {\sf{v \: = \: 0 \: + \: 10 \: \times \: 3}} \\ \\ \\ \implies {\sf{v \: = \: 30}} \\ \\ \\ \large {\boxed{\sf{Final \: Velocity \: at \: 3s \: is\: 30 \: ms^{-1}}}}$

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Use formula :

$\large \bigstar {\boxed{\sf{v^2 \: = \: u^2 \: + \: 2gh}}} \\ \\ \\ \implies {\sf{v^2 \: = \: (0)^2 \: + \: 2(10)(100) \\ \\ \\ \implies {\sf{v^2 \: = \: 2000}} \\ \\ \\ \implies {\sf{v \: = \: \sqrt{2000}}} \\ \\ \\ \implies {\sf{v \: = \: 44.7}}$

Final Velocity = 44.7 m/s

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Use formula :

$\large \bigstar {\boxed{\sf{s \: = \: ut \: + \: \dfrac{1}{2}gt^2}}} \\ \\ \\ \implies {\sf{100 \: = \: \dfrac{1}{2}(10)(t)^2}} \\ \\ \\ \implies {\sf{t^2 \: = \: \dfrac{100 \: \times \: 2}{10}}} \\ \\ \\ \implies {\sf{t^2 \: = \: \dfrac{200}{10}}} \\ \\ \\ \implies {\sf{t^2 \: = \: 20}} \\ \\ \\ \implies {\sf{t \: = \: \sqrt{20}}} \\ \\ \\ \implies {\sf{t \: = \: 4.47}}$

Time is 4.47 s