Question

A cricket ball is dropped from a height of 20 m.

$\bullet$ Calculate the Speed of ball when it hits the ground.

$\bullet$ Calculate the time it takes to fall through this Height (g = 10 m/s²)​

Answer 1

Solution :-

Given :-

• Height (h) = 20 m
• Initial Velocity (u) = 0 m/s
• Acceleration due to gravity (g) = 10 N

To Find :-

• Speed of ball when it hits the ground (v) i.e Final Velocity
• Time Taken

Calculating Final Velocity:-

Formula Used :-

• v² - u² = 2gh

Substituting values and solving further :-

⟹ v² - 0² = 2 × 10 × 20

⟹ v² = 400

⟹ v = $\sf \sqrt{400}$

⟹ v = 20

Therefore, the final velocity is 20 m/s

Now,

Calculating Time Taken :-

Formula Used :-

• v = u + gt

Substituting values and solving further :-

⟹ 20 = 0 + 10 × t

⟹ 20 = 10t

⟹ 10t = 20

⟹ t = 20/10

⟹ t = 2

Therefore, time taken is 2 seconds.

Important Formulas :-

• v² - u² = 2gh
• v = u + gt
• h = ut + ½gt²

Answer 2

Given Question:

A cricket ball is dropped from a height of 20 m.

• Calculate the Speed of ball when it hits the ground.

• Calculate the time it takes to fall through this Height (g = 10 m/s²)

How to solve?

As per it is given in the question that a ball is dropped from the height so its initial velocity represented by 'u' will be 0m/s. And height is also given which is 20m.

To find the speed which will be final velocity of the ball, we'll use the formula ||v²= u² + 2gh||, where, v is the final velocity, u is initial velocity and g is acceleration due to gravity and h is height.

And,

To find the time we'll use the formula ||v =u + gt||. Where, t is time, v is the final velocity, u is initial velocity and g is acceleration due to gravity which is 10m/s.

Required Solution:

$\huge \tt { \underline { Given \: Information -}}$

• $\sf { ( u )\: Initial \:velocity = 0m/s}$

• $\sf { ( h )\: Height= 20m}$

• $\sf { ( g )\: acceleration \: due \: to \: gravity= 10m/{s} ^{2} }$

Calculation:

Calculate the Speed of ball when it hits the ground.

We know that,

• ${\underline {\underline {\rm {\red { {v} ^{2 } ={u} ^{2 } + 2gh}}}}}$

Therefore,

$\large \tt {\qquad {v} ^{2 } ={u} ^{2 } + 2gh}$

$\large \tt {\qquad \implies {v} ^{2 } ={0} ^{2 } + 2 \times 10 \times 20 }$

$\large \tt {\qquad \implies {v} ^{2 } ={0} ^{2 } + 20 \times 20 }$

$\large \tt {\qquad \implies {v} ^{2 } ={0} ^{2 } + 400 }$

$\large \tt {\qquad \implies {v} ^{2 } = 400 }$

$\large \tt {\qquad \implies v= \sqrt{400} }$

$\large \tt\purple {\qquad \implies v= 20m/s }$

$\therefore$ Final velocity of the ball is 20m/s.

Now, Calculate the time it takes to fall through this Height (g = 10 m/s²)

We know that,

• ${\underline {\underline {\rm {\red { v = u + gt} }}}}$

Therfore,

$\large \tt {\qquad v = u + gt}$

$\large \tt {\qquad \implies 20 = 0 + 10 \times t}$

$\large \tt {\qquad \implies 20 = 0 + 10t}$

$\large \tt {\qquad \implies 20 = 10t}$

$\large \tt {\qquad \implies t = \cancel { \dfrac{20}{10}} =2sec}$

$\therefore$ The time it takes to fall through this height is 2sec.

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Points to remember-

• Whenever the body is thrown vertically upwards, then ( u ) initial velocity is 0.

• Whenever the body is thrown vertically upwards then ( v ) final velocity is 0.

Normal equations of motion:

• $\rm { v = u + at}$

• $\rm { {v} ^{2 } ={u} ^{2 } + 2as }$

• $\rm { s = ut + \dfrac{1}{2} a{t} ^{2}}$

Equations of motion for freely falling bodies -

• $\rm { v = u + gt}$
• $\rm { {v} ^{2 } ={u} ^{2 } + 2gh }$
• $\rm { s = ut + \dfrac{1}{2} g{t} ^{2}}$

Only the difference is that in equations of motion for freely falling bodies, s ( displacement /distance) changes to height and a ( acceleration) changes to g ( acceleration due to gravity).

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