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solmaris [256]
2 years ago
11

if you drop a stone from height of 2.5m. what is the speed of the stone right before it hits the ground?

Physics
1 answer:
KonstantinChe [14]2 years ago
6 0
Since the stone was dropped from height, its initial velocity = 0 m/s

Using  v² = u² + 2gs.

Where g ≈ 10 m/s²,  u = initial velocity = 0 m/s, s = height from drop = 2.5 m

v² = u² + 2gs

v² = 0² + 2*10*2.5

v² = 0 + 50

v² = 50

v = √50

v ≈ 7.07 m/s

Hence velocity just before hitting the ground is ≈ 7.07 m/s 
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From the question we are told that:

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Therefore

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 W_m=183.495N

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A hiker walks from
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(a) The distance travelled by the hiker is 7.16 km

(b) If the hiker desires to return to his starting point, he should move at 201.71° with respect to east.

Form a right angled triangle with the distance between initial and final points as hypotenuse. The adjacent side will be the total distance travelled on x-direction and the opposite side will be the total distance travelled in y-direction.

Adjacent side = x_{2} - x_{1} = 3.05 - ( - 3.6 )

Adjacent side = 6.65 km

Opposite side = y_{2} - y_{1} = 5.9 - 3.25

Opposite side = 2.65 km

According to Pythagoras theorem,

Hypotenuse^{2} = Opposite side^{2} + Adjacent side^{2}

Hypotenuse^{2} = 2.65^{2} + 6.65^{2}

Hypotenuse = √ ( 7.02 + 44.22 )

Hypotenuse = 7.16 km

sin θ = Opposite side / Hypotenuse

sin θ = 2.65 / 7.16

θ = sin^{-1} ( 0.37 )

θ = 21.71°

This is the angle in which the hiker travelled from initial to final direction. The angle in which the hiker should travel to return to initial point with respect to east is

θ = 21.71° + 180°

θ = 201.71°

Therefore,

(a) The distance travelled by the hiker is 7.16 km

(b) If the hiker desires to return to his starting point, he should move at 201.71° with respect to east.

To know more about Pythagoras theorem

brainly.com/question/343682

#SPJ1

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