Question

→ 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

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.

━━━━━━━━━━━━━━━━━━

Answer 2

Answer:

v²-u² = 2gs

(0)²-(40)² = 2 × 10 × s

- 1600 = 20 × s

S = -1600/20

s = -80 m