Question

A ball is dropped from a height of 2.5 m. find the velocity when it reaches the floor?​

Answer 1

Given :

• Height of building, h = 2.5 m
• Initial velocity, u = 0 m/s [Dropped from height]
• Acceleration due to gravity, g = 9.8 m/s² [Down]

To Find :

• Final velocity, v

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By using the 3rd equation of motion for fréély falling bodies,

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

• v denotes final velocity
• u denotes initial velocity
• g denotes acceleration due to gravity
• h denotes height

$\quad \twoheadrightarrow\sf { v^2 - (0)^2 = 2 \times 9.8 \times 2.5}$

$\quad \twoheadrightarrow\sf { v^2= 19.6 \times 2.5}$

$\quad \twoheadrightarrow\sf { v^2= 49}$

$\quad \twoheadrightarrow\sf { v = \sqrt{49}}$

$\quad \twoheadrightarrow\boxed{\sf { v = 7 \; ms^{-1} }}$

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Equations of motion for fréély falling bodies,

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

Here,

• v denotes final velocity
• u denotes initial velocity
• t denotes time
• g denotes acceleration due to gravity
• h denotes height

Answer 2

Exρlαnαtion:

ㅤㅤㅤㅤㅤㅤㅤIt is given that : A ball is dropped from a height of 2.5 m . and We are required to find the final velocity when it reached floor.

As it dropped Initial velocity (u = 0 m/s) and h = 2.5m v=?

Now, by using equations of motion of freely falling body we can find the final velocity.

v² - u² = 2gh

>> v = final velocity

>> u = initial velocity

>> g = acceleration due to gravity (9.8 m/s²)

>> h = height

Substituting the values,

>> v² - (0) = 2 (9.8)(2.5)

>> v² = 2 × (98×25)/100

>> v² = 2 × (98/4)

>> v² = 98/2

>> v² = 49

v = √49

${\boxed{\sf\large\orangev = 7 m/s}$

So, the velocity when it reaches the floor is 7 m/s.

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