Question

It a physics question

Answer 1

$\huge\mathfrak\red{Answer :}$


According to question we have given;

Distance (s) = 20 m

initial velocity (u) = 0

Acceleration (a) = 10 m/s²

final velocity (v) = ?

Formula used here is...

v² - u² = 2as

v² = 2as + u²

v² = 2(10)(20) + (0)²

v² = 400

v = √400

$\textbf{v = 20 m/s}$

Also we have to find time (t)...

Now,

v = u + at

We know the values... just put them...

20 = 0 + 10 t

10 t = 20

t = $\dfrac{20}{10}$

$\textbf{t = 2 sec.}$

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Answer 2

Hey.


Here is the answer.


Given : Dropped from the height, s = 20 m

velocity at the time of dropping, u = 0 m/s

{as it is gently dropped}

acceleration = 10 m/s^2


As we know that

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

so, velocity with which it will hit the ground
will be

$v \: = \: \sqrt{2as \: } \: \: \: \: \: \: as \: u \: = \: 0 \: m {s}^{ - 1} \\ \\ so \: \: v \: = \: \sqrt{2 \times 10 \times 20} \\ \\ so \: \: v \: = \: 20 \: m {s}^{ - 1}$


So it will hit the ground with a velocity of 20m/s


Also, we know that

$t \: = \: \frac{v - u}{a} \\ \\ as \: \: v \: = \: u \: + \: at$


so, time to reach the ground will be

$t \: = \: \frac{20 - 0}{10} \\ \\ or \: \: t \: = \: 2 \: s$


So it will reach the ground after 2s after dropping gently from the height 20m.


Thanks