Question

A ball falls off a table and reaches the ground in 1s. assuming g=10m/s2 calculate it's speed on reaching the ground and the height of the table

Answer 1

v=gt
=10×1
=10m/s
$v = gt = \sqrt{2gh}$
(10)^2=2×10×h
100=20h
h=100/20=5m

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Answer 2

Answer:

The height of the table is 5 m.

The velocity of the ball on reaching the ground is 10 m/s.

Explanation:

Given data:

$\bullet\sf\:Time\:(\:t\:)\:=\:1\:s\\\\\\\bullet\sf\:Acceleration\:due\:to\:gravity\:(\:g\:)\:=\:10\:m\:/\:s^2\\\\\\\bullet\sf\:Initial\:velocity\:(\:u\:)\:=\:0\:m\:/\:s$

To find:

$\sf\:1\:)\:Height\:of\:table\:(\:s\:)\\\\\\\sf\:2\:)\:Speed\:of\:ball\:(\:v\:)$

Solution:

By using Newton's second equation of motion for freely falling objects,

$\pink{\sf\:s\:=\:\dfrac{1}{2}\:g\:t^2}\\\\\\\implies\sf\:s\:=\:\dfrac{1}{\cancel2}\:\times\:\cancel{10}\:\times\:1\\\\\\\implies\sf\:s\:=\:5\:\times\:1\\\\\\\implies\boxed{\red{\sf\:s\:=\:5\:m\:}}$

The height of the table is 5 m.

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Now,

By using Newton's first equation of motion for freely falling objects,

$\pink{\sf\:v\:=\:u\:+\:g\:t}\\\\\\\implies\sf\:v\:=\:0\:+\:10\:\times\:1\\\\\\\implies\sf\:v\:=\:0\:+\:10\\\\\\\implies\boxed{\red{\sf\:v\:=\:10\:m\:/\:s}}$

The velocity of the ball on reaching the ground is 10 m / s.

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Additional Information:

1. Free fall:

When an object falls on the ground only because of the gravity of earth, it is called a freely falling object and the process known as free fall.

2. Newton's Equations of motion for Freely falling objects:

$\pink{\sf\:1.\:v\:=\:g\:t}\\\\\\\red{\sf\:2.\:s\:=\:\dfrac{1}{2}\:g\:t^2}\\\\\\\green{\sf\:3.\:v^2\:=\:2\:g\:s}\\\\\\\sf\:Where\:,\\\\\\\longrightarrow\sf\:v\:=\:Final\:velocity\\\\\\\longrightarrow\sf\:g\:=\:Acceleration\:due\:to\:gravity\\\\\\\longrightarrow\sf\:t\:=\:Time\\\\\\\longrightarrow\sf\:s\:=\:Distance$