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?

Answer 1

Answer:

Initial Velocity u=40

Fianl velocity v=0

Height, s=?

By third equation of motion

v

2

−u

2

=2gs

0−40

2

=−2×10×s

s=

20

160

⇒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 .

Answer 2

Answer :-

i) Maximum height  =80m

ii) Total distance = 160m

iii) Net displacement = 0

Explanation :-

Given :

Initial velocity = 40m/s

Gravitational acceleration,g = 10m/s

Final velocity = 0m/s      [as stone was at its maximum height]

To Find :

Maximum height reached by the stone = ?

Net displacement = ?

Total distance = ?

Solution :

Accirding to the first equation of motion,

$\boxed{\sf{}v=u+at}$         [here h is same as g]

Where,

u is the initial velocity of the stone.

v is the final velocity of the stone.

s is the height of the stone.

g is the acceleration due to gravity.

Let h be the maximum height attained by the stone.

Therefore,

$\sf{}:\implies 0 - (40)^2 = 2 \times h\times (-10)$

$\sf{}:\implies 0 - 1600 = -20 \times h$

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

$\sf{}\therefore h =80m$

Hence,maximum height attained by the stone is 80m

Total distance covered by the stone during its upward and downward journey = (80 + 80)m

= 160m

Therefore,total distance travelled by the stone is equal to 160m

Displacement is the shortest distance from initial to the final position

Net displacement of the stone during its upward and downward journey

= 80 + (−80)

= 0

Therefore,displacement is equal to zero.