Physics, asked by ObnoxiousS, 18 days ago

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?​

Answers

Answered by xxRoyalgirlxx
4

Answer:

Initial Velocity u=40

Fianl velocity v=0

Height, s=?

By third equation of motion

v²-u²=2gs

0−40²=−2×10×s

s= 160/20

⇒s=80m/s

Toatl distance travelled by stone = upward distance + downwars distance =2×s=160m

Total Diaplacement =0, Since the initial and final point is same.

Answered by StarFighter
6

Given :-

  • A stone is thrown vertically upward with an initial velocity of 40 m/s.
  • Acceleration due to gravity (g) = 10 m/s².

To Find :-

  • What is the maximum height reached by the stone.
  • What is the net displacement.
  • What is the total distance covered by the stone.

Formula Used :-

\clubsuit Third Equation Of Motion Formula :

\bigstar \: \: \sf\boxed{\bold{\pink{v^2 =\: u^2 + 2gh}}}\: \: \: \bigstar\\

where,

  • v = Final Velocity
  • u = Initial Velocity
  • g = Acceleration due to gravity
  • h = Height

Solution :-

\sf\bold{\underline{\purple{\dag\: (i)\: The\: maximum\: height\: reached\: by\: the\: stone\: :-}}}\\

Given :-

  • Final Velocity (v) = 0 m/s [as stone was at its maximum height]
  • Acceleration due to gravity (g) = 10 m/
  • Initial Velocity (u) = 40 m/s

According to the question by using the formula we get,

\implies \bf v^2 =\: u^2 + 2gh\\

\implies \sf (0)^2 =\: (40)^2 + 2(- 10)h

\implies \sf (0 \times 0) =\: (40 \times 40) + (- 20)h

\implies \sf 0 =\: 1600 - 20h

\implies \sf 0 - 1600 =\: - 20h

\implies \sf {\cancel{-}} 1600 =\: {\cancel{-}} 20h

\implies \sf 1600 =\: 20h

\implies \sf \dfrac{1600}{20} =\: h

\implies \sf 80 =\: h

\implies \sf\bold{\purple{h =\: 80\: m}}\\

\therefore The maximum height reached by the stone is 80 m .

\sf\bold{\underline{\blue{\dag\: (ii)\: Net\: Displacement\: :-}}}\\

As we know that :

\leadsto Displacement is the shortest distance between the initial and final velocity position of a moving object in particular direction.

So, the net displacement of the stone during it upward and downward :

\dashrightarrow \sf 80 + (- 80)

\dashrightarrow \sf 80 - 80

\dashrightarrow \sf\bold{\blue{0}}\\

\therefore The net displacement of the stone is zero.

\small \sf\bold{\underline{\red{\dag\: (iii)\: Total\: Distance\: Covered\: By\: The\: Stone\: :-}}}\\

\leadsto Total distance covered by the stone is :

\longrightarrow \sf (80 + 80)\: m

\longrightarrow \sf\bold{\red{160\: m}}\\

\therefore The total distance covered by the stone is 160 m .

Similar questions