Question

a stone is dropped from a height of 40 m. a) how much time will it take to reach the ground. b) with what velocity will it strike the ground​

Answer 1

Answer

Given -

$\longrightarrow$u = 0 ( because stone is dropped )

$\longrightarrow$s = 40m

$\longrightarrow$a = g = 9.8 m/s²

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To find -

a) Final velocity $\longrightarrow$ v

b) Time $\longrightarrow$ t

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Formula used -

1st equation of motion -

$\longrightarrow$v = u + at

3rd equation of motion -

$\longrightarrow$v² = u² + 2as

where v is final velocity , u is initial velocity, a is acceleration due to gravity and s is distance travelled.

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Solution -

By 3rd equation of motion -

v² = u² + 2as

$\longrightarrow$v² = 0 + 2 × 9.8 × 40 m/s

$\longrightarrow$v² = 784

v = 28

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Using 1st equation of motion -

v = u + at

$\longrightarrow$28 = 0 + 9.8 t

$\longrightarrow$t = 28/9.8

$\longrightarrow$t = 2.8 s

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ADDITIONAL INFORMATION

Acceleration is the rate of change of velocity with respect to time.

Equation of motion are used only when uniform acceleration acts on body.

$\longrightarrow$s = ut + 1/2 gt²

$\longrightarrow$Snth = u + a/2 ( 2n - 1 )

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Thanks

Answer 2

✧ Correct Question:

☞ A stone is dropped from a height of 40 m, then find , How much time will it take to reach the ground h what velocity will it strike the ground.

✧ To Find:

• Time taken to reach the ground.

• Final velocity of the stone.

✧ Given:

• Height = 40 m
• g = $10 ms^{-2}$

✧ We know:

• Second Law of motion :

$\boxed{\sf{h = ut \pm \dfrac{1}{2}gt^{2}}}$

☞ where,

• h = height
• u = intial velocity
• g = acceleration due to gravity
• t = time taken

• Third law of motion :

$\boxed{\sf{v^{2} = u^{2} - 2gh}}$

☞ where,

• v = Final velocity
• u = initial Velocity
• g = acceleration due to gravity
• h = height

✧ Concept:

☞ In this case , the the initial velocity will be zero , and the since it is with the gravity the , the gravity will be positive.

✧Solution:

• For finding the time taken to reach the ground:-

☞ we know ,

$\it{h = ut + \dfrac{1}{2}gt^{2}}$

$\because$ It is with the gravity.

☞ Putting the value in the equation , we get :-

$\mathtt{\Rightarrow 40 = 0 \times t + \dfrac{1}{2} \times 10 \times t^{2}}$

$\mathtt{\Rightarrow 40 = \dfrac{1}{\cancel{2}} \times \cancel{10} \times t^{2}}$

$\mathtt{\Rightarrow 40 = 5 \times t^{2}}$

$\mathtt{\Rightarrow \dfrac{40}{5} = t^{2}}$

$\mathtt{\Rightarrow \cancel{\dfrac{40}{5}} = t^{2}}$

$\mathtt{\Rightarrow 8 = t^{2}}$

$\mathtt{\Rightarrow \sqrt{8} = t}$

$\mathtt{\Rightarrow 2\sqrt{2} s = t}$

$\boxed{\therefore time\:taken = 2\sqrt{2} s = t}$

• For Finding the Final velocity:-

☞ we know ,

$\sf{v^{2} = u^{2} + 2gh}$

☞ Putting the value in the equation , we get :-

$\mathtt{\Rightarrow v^{2} = 0^{2} + 2 \times 10 \times 40}$

$\mathtt{\Rightarrow v^{2} = 0^{2} + 2 \times 10 \times 40}$

$\mathtt{\Rightarrow v^{2} = 800}$

$\mathtt{\Rightarrow v = \sqrt{800} ms^{-2}}$

$\mathtt{\Rightarrow v = 28.28 ms^{-2}}$

$\boxed{\therefore v = 28.28 ms^{-2}}$

Extra Information:

• $\sf{h_{max} = \dfrac{u^{2}}{2g}}$ [v = 0]

• $\it{h = \dfrac{1}{2}gt^{2}}$ [u = 0]

• $\it{v = u + at}$

• $\it{s_{n} = u + \dfrac{1}{2}a(2n - 1)}$

where,

• h = height
• v = final velocity
• u = initial velocity
• g = acceleration due to gravity
• a = acceleration
• t = time taken