Question

a stone is dropped from a height of 20m . 1》find the speed when it hits the ground ?​

Answer 1

here is the answer

Explanation:

h=20m/s

u=0

v=?

t=?

v²-u²=2gh

v²-o²=2*10m/s * 20m

v=√400

v=20m/s[velocity with which stone hit the ground]

v=u+gt

20=0+10*t

t=2s[time to reach the ground]

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Answer 2

Solution :-

♦ As given :-

• The height from which the stone is dropped = 20 m

♦Now as the stone is falling under the influence of gravitational pull of earth.

♦We will take

• Height as positive

• a = g = 9.8 m/s²

♦Now we will find out time first

→S = ut + 1/2 at²

=> 20 = 0(t) + 1/2 × 9.8 × t²

=> 20 = 4.9 t²

$\implies t^2 = \dfrac{20}{4.9}$

$\implies t^2 = 4.082$

$\implies t = \sqrt{4.082}$

$\implies t = 2.020$

$\implies t = 2.02 \: sec$

Now as t = 2.02

♦Then from motion equation

→ v = u + at

=> v = 0 + 9.8 × 2.02

=> v = 19.796

=> v = 19.8 m/s (in the direction of gravity)

Or

=> v = -19.8 m/s in the direction opposite of gravity.

So it will hit ground with the velocity of 19.8 m/s .