Question

a ball is dropped from a height of 20m if its velocity increases uniformly at the rate of 10ms^2 with what velocity will it strike the ground? ​

Answer 1

Answer:

Given :

a = 10 m/s²

s = 20 m

u = 0 m/s

So applying third equation of motion :

v² - u² = 2as

⇒ v² - 0² = 2 × 10 × 20

⇒ v² = 400

⇒ v = √400

⇒ v = 20 m/s

Hence, the velocity at which the ball will strike the ground is 20 m/s.

Some Important Formulas :

• v = u + at
• v² - u² = 2as
• s = ut + ½at²
• Average velocity = ∆x/∆t
• Average Speed = Total Path/Total Time
• Instantaneous velocity = dx/dt
• Averages acceleration = ∆v/∆t
• Instantaneous acceleration = dv/dt
• ∆x = x_2 - x_1

Answer 2

Given :-

• A ball is dropped from a height of 20 m if it's velocity increases uniformly at the rate of 10 ms²

To Find :-

• Velocity will it strike the ground

Solution :-

We have,

• Acceleration(a) = 10 m/s²
• Height(s) = 20 m
• Initial Velocity(u) = 0 m/s

We know that,

$\boxed{\sf v^{2} - u^{2} = 2as} \: \: (\bf 3rd \: equation \: of \: motion)$

Where,

• v = Final Velocity
• u = Initial Velocity
• a = Acceleration
• s = Distance Covered

Substituting the given values we get,

$:\implies\sf v^{2} - 0^{2} = 2 \times 10 \times 20$

$:\implies\sf v^{2} - 0^{2} = 400$

$:\implies\sf v^{2} = 400$

$:\implies\sf v = \sqrt{400}$

$:\implies\underline{\boxed{\blue{\sf v = 20\:m/s}}}$

Hence, the velocity will it strike the ground is 20 m/s.

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Additional Information :-

★ 1st equation of motion :-

$\boxed{\sf v = u + at}$

Where,

• v = Final Velocity
• u = Initial Velocity
• a = Acceleration
• t = Time taken

★ 2nd equation of motion :-

$\boxed{\sf s = ut + \frac{1}{2} at^{2}}$

Where,

• s = Distance Covered
• u = Initial Velocity
• t = Time taken

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