Question

A stone is dropped from the height of 121m and reaches to the earth in 10 sec with an acceleration of 20m/s square find the initial velocity of the stone​

Answer 1

Given

• Height = 121 m
• Time = 10 sec
• Acceleration = 20 m/s²

To Find

• Initial Velocity of the stone

Solution

☯ v²-u² = 2as

• Here we shall simply use the third equation of motion to find the initial velocity as we have all the other values.

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✭ According to the Question :

➞ v²-u² = 2as

• v = Final Velocity = 0 m/s
• u = Initial Velocity = ?
• a = Acceleration = 20 m/s²
• s = Distance = 121 m

➞ 0² - u² = 2 × 20 × 121

➞ -u² = 4840

➞ -u = √4840

➞ u = - 69.5 m/s

∴ The initial velocity of the body is -69.5 m/s

Answer 2

Answer:

Given :-

• Height = 121 m
• Time = 10 sec
• Acceleration = 20 m/s²

To Find :-

Initial Velocity

Solution :-

As we know that

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

V = Final Velocity

U = Initial Velocity

A = Acceleration

S = Distance

0² - u² = 2(20)(121)

0 - u² = 4840

-u² = 4840

-u = √4840

u = -69.5 m/s