Question

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→ A stone is thrown vertically upward with an initial velocity of 40 m/s. Taking g = 10 m/s², find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone when it falls back to the ground ?​

Answer 1

Answer:

• Maximum height attained by the stone = 80m
• Net Displacement = 0 m
• Total Distance covered by stone = 160 m

Explanation:

Given:-

• Initial velocity ,u = 40m/s
• Final velocity ,v = 0m/s (as the stone reached its maximum height)
• Acceleration due to gravity ,g = 10m/s²

To Find:-

• Maximum height ,h
• Net Displacement ,S
• Total distance covered by ball , d

Solution:-

Firstly we calculate the maximum height attained by the stone when it is thrown . Using 3rd equation of motion

• v² = u² + 2gh

where,

• v denote final velocity
• u denote initial velocity
• g denote acceleration due to gravity
• h denote maximum height attained by stone

Substitute the value we get

→ 0² = 40² + 2 (-10) × h

→ 0 = 1600 -20h

→ -1600 = -20h

→ 20h = 1600

→ h = 1600/20

→ h = 80 m

• Hence, the maximum height attained by the stone is 80 metres

The total distance covered by the stone when it falls back to the ground is 80 + 80 = 160 metres

And the net displacement of the stone is 0 metres (as the stone reached its initial position).

Answer 2

Concept:- Free Fall - Motion of Objects Under the Influence of Gravitational Force of the Earth

$\bf \: Initial \: velocity \: u = 40$

$\bf \: Final \: velocity \: v = 0$

$\bf \: Height, s = \: ?$

$\bf \: By \: third \: equation \: of \: motion$

$\bf \: v²-u²=2gs$

$\bf 0=40²=-2×10×s$

$\bf \: s = \frac{ \: 160 \: }{ \: 20 \: }$

$\bf \: s =80m/s$

Total distance travelled by stone = upward distance + downward distance

$\bf2×s=160m$

$\bf \: Total \: Displacement=0$

Since, the initial and final point is same.