→ A stone is thrown vertically upward with an initial velocity of 40

1.	→ 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» Given:- Initial Velocity {u} = 40 m/s Final Velocity {v} = 0 To Find:- Height {s} Total Displacement Equation USED:- v² - u² = 2gs {Third equation of motion} SOLUTION:- ➠ v² - u² = 2gs ➠ (0)² - (40)² = 2 × 10 × s ➠ - 1600 = 20 × s ➠ s = $\dfrac{-1600}{20}$ ➠ s = -80 m Height {s} can't be negative So, s = 80 m. Total distance travelled by stone = Upward distance + downward distance ⠀⠀⠀⠀= 2 × s ⠀⠀⠀⠀= 2 × 80 ⠀⠀⠀⠀= 160 m As initial velocity and final velocity is same Total Displacement = 0 Hence, the Height and Total Displacement are 80 m and 0 m respectively. ━━━━━━━━━━━━━━━━━━

→ 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»

➠ v² - u² = 2gs

➠ (0)² - (40)² = 2 × 10 × s

➠ - 1600 = 20 × s

➠ s = $\dfrac{-1600}{20}$

➠ s = -80 m

Height {s} can't be negative So, s = 80 m.

Total distance travelled by stone = Upward distance + downward distance

⠀⠀⠀⠀= 2 × s

⠀⠀⠀⠀= 2 × 80

⠀⠀⠀⠀= 160 m

As initial velocity and final velocity is same Total Displacement = 0

Hence, the Height and Total Displacement are 80 m and 0 m respectively.

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