Question

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Answer 1

Answer:

Step-by-step explanation:

3)

$\sf \boxed{v_{f}^{2}=v_{i}^{2}+2gd_{y}}\\\\v_{f} - > final \ velocity\\\\v_{i} - > Initial \ velocity = 0 \\\\d_{y} - > distance \ travelled$

g -> gravitational acceleration = 9.8 m/s²

$v_{f}^{2}= 0 + 2*9.8 * 45$

    = 882 m

Answer 2

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Answer 1 :-

Here, We have

• Initial velocity = 0 , as jeepney started from rest
• Time = 2 seconds
• Distance = 20 m

Therefore,

By using second equation of motion

$\sf{ S = ut {\times} }{\sf{\dfrac{1}{2}}}{\sf{at^{2}}}$

Subsitute the required values,

$\sf{ 20 = 0{\times}2 {\times} }{\sf{\dfrac{1}{2}}}{\sf{{\times}a(2)^{2}}}$

$\sf{ 20 = }{\sf{\dfrac{1}{2}}}{\sf{{\times}4a}}$

$\sf{ 20 = 2a }$

$\sf{ a = }{\dfrac{ 20}{2}}$

$\sf{ a = 10\:m/s^{2}}$

Hence, The acceleration of the jeepeny is 10 m/s² .

Answer 2 :-

Here, we have

• Initial velocity = 0 , as the coins started rolling from rest.
• Acceleration = 4 m/s²
• Time = 3 seconds

Therefore,

By using second equation of motion

$\sf{ S = ut {\times} }{\sf{\dfrac{1}{2}}}{\sf{at^{2}}}$

Subsitute the required values,

$\sf{ S = 0{\times}3 {\times} }{\sf{\dfrac{1}{2}}}{\sf{{\times}4(3)^{2}}}$

$\sf{ S = }{\sf{\dfrac{1}{2}}}{\sf{{\times}4(9)}}$

$\sf{ S = }{\sf{\dfrac{1}{2}}}{\sf{{\times}36}}$

$\sf{ S = 18 \: m }$

Hence, The coin will reach 18 m after 3 seconds.

Answer 3 :-

Here,

• A stone fell from a 45 m high cliff and hit the ground.

So,

• Initial velocity = 0
• Distance = 45 m
• Acceleration due to gravity

By using third equation of motion

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

Subsitute the required values,

$\sf{ v^{2} = 0^{2} + 2{\times}9.8{\times}45 }$

$\sf{ v^{2} = 90{\times}9.8}$

$\sf{ v^{2} = 882 m}$

$\sf{ v = \sqrt{882} m}$

$\sf{ v = 29.7 m/s}$

Hence, The final velocity of the stone when the stone hits the ground is 29.7 m.