Question

$\bf{Question}$
A stone is thrown vertically upward with an initial velocity of 40 m/s. Taking g= 10 m/s, find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone?

Note-Well defined answer needed!!​

Answer 1

The maximum height is 80 m and the total distance covered by the stone is 160 m and the displacement is zero.

Explanation:

Given that,

Final velocity v = 0

Initial velocity u = 40m/s

We know that,

Using equation of motion

v^2=u^2+2ghv

2

=u

2

+2gh

0-40^2= 2\times 10\times h0−40

2

=2×10×h

The maximum height is

h = 80\ mh=80 m

The stone will reach at the top and will come down

Therefore, the total distance will be

s = h_{1}+h_{2}s=h

1

+h

2

s = 80\ m+80\ m = 160 ms=80 m+80 m=160m

The net displacement is

D = h_{1}-h_{2}D=h

1

−h

2

D = 80\ m-80\ m= 0D=80 m−80 m=0

Hence, The maximum height is 80 m and the total distance covered by the stone is 160 m and the displacement is zero.

Answer 2

Answer:

Given :-

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

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 :

$\longrightarrow \sf\boxed{\bold{\pink{v^2 =\: u^2 + 2as}}}$

where,

• v = Final Velocity
• u = Initial Velocity
• a = Acceleration
• s = Distance Covered

Solution :-

★ Maximum height reached by the stone :-

Given :

• Initial Velocity = 40 m/s
• Acceleration due to gravity = - 10 m/s [As the object is thrown vertically upwards]
• Final Velocity = 0 m/s

According to the question by using the formula we get,

$\implies \sf (0)^2 =\: (40)^2 + 2(- 10) \times s$

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

$\implies \sf 0 =\: 1600 - 20 \times s$

$\implies \sf 0 - 1600 =\: - 20s$

$\implies \sf {\cancel{-}} 1600 =\: {\cancel{-}} 20s$

$\implies \sf 1600 =\: 20s$

$\implies \sf \dfrac{1600}{20} =\: s$

$\implies \sf 80 =\: s$

$\implies \sf\bold{\red{s =\: 80\: m}}$

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

________________

★ Net Displacement :-

$\leadsto$ Now, when we throw stone upwards its reaches highest point and come down.

So, the initial point and final point is same.

Hence, the total displacement is zero.

$\therefore$ Net Displacement = 0

________________

★ Total Distance Covered by the stone :-

$\footnotesize \dashrightarrow \bf Total\: Distance\: Covered =\: 2 \times Maximum\: Height\: Covered\: by\: stone$

$\dashrightarrow \sf Total\: Distance\: Covered =\: 2 \times 80$

$\dashrightarrow \sf\bold{\red{Total\: Distance\: Covered =\: 160\: m}}$

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

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