Question

A ball is gently dropped from a height of 20m .id its velocity increases uniformly at the rate of 10ms ,with what velocity will it strike the ground ?after what time will it strike the ground?

Answer 1

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✓✓A ball is gently dropped from a height of 20m. If it's velocity increases uniformly at the rate of 10m/s² ,with what velocity will it strike the ground? After what time will it strike the ground?

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• $\bf\:Initial\: velocity\:(u)= 0\:m/s$

• $\bf\:Distance\:(s)= 20\:m$

• $\bf\:acceleration\:(a)=10m/{s}^{2}$

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• $\bf\:(i)\:Final\: velocity\:(v)=\:?$

• $\bf\:(ii)\:Time\:taken\:(t)=\:?$

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$\bf\:{v}^{2}={u}^{2}+2as$

$\bf\implies\:{v}^{2}={0}^{2}+2\times\:10\times\:20$

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

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

$\bf\implies\:v=20\:m/s$

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$\bf\:v=u+at$

$\bf\therefore\:t=\dfrac{v-u}{a}$

$\bf\implies\:t=\dfrac{20-0}{10}$

$\bf\implies\:t=\dfrac{20}{10}$

$\bf\implies\:t=2\:sec$

Therefore,

Final velocity will be 20 m/s which the ball will strike the ground.

and time taken will be 2 sec after which the ball will strike the ground.

Answer 2

Answer:

After 2 sec

Explanation:

Initial velocity = 0 m/s, height = 20 m, acceleration = 10 m/s^2

Time = ?

v^2 - u^2 = 2as (Third Equation Of Motion)

v^2 - 0^2 = 2*10*20

v^2 = 400

v = 20 m/s

v = u + at (First Equation Of Motion)

20 = 0 + 10t

10t = 20

t = 2 sec