Question

A boy standing at the top of a 20-metre-high tower drops a stone. Assuming g = 10 , the velocity

with which the stone hits the ground is​

Answer 1

Answer:

velocity with which the stone hits the ground is 20 m/s

Explanation:

Using eauations of motion:

• v^2 = u^2 + 2aS

Here,

As the stone is dropped, its initial velocity is 0. It's going vertically downward, acceleration is gravitational acceleration of g.

Here, the final velocity be v

= > v^2 = 0^2 + 2aS

= > v^2 = 2(10)(20)

= > v^2 = 400

= > v = √(400) = 20 m/s

Answer 2

Answer:

⇢ ${\boxed{\bold{Velocity = 20 \: m/s}}}$

Explanation:

Given:-

⇢ A boy standing at the top of a 20-metre-high tower drops a stone. Assuming g = 10.

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Find:-

⇢ The velocity with which the stone hits the ground is.

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Using formula:-

★ ${\boxed{\bold{v^{2} = u^{2} + 2as}}}$

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Know terms:-

• Velocity = (v).
• Initial velocity = (u).
• Acceleration = (a).
• Displacement = (s).

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Calculations:-

• Initial velocity is also '0' as it's is the starting velocity.

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⇢ $\rm{v^{2} = 0^{2} + 2as}$

⇢ $\rm{v^{2} = 2(10 \times 20)}$

⇢ $\rm{v^{2} = 400}$

⇢ $\rm{v = \sqrt{400}}$

⇢ ${\boxed{\bold{v = 20 \: m/s}}}$

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Therefore, 20 m/s is the required velocity.