Question

A stone thrown vertically upwards travels to a height 6 m and falls
back to the hand. On the basis of the above statement answer the question
1. When the stone reaches the topmost position what is the distance travelled by the stone. And what is the displacement of the stone
2. When the stone falls back in the hand what is the distance travelled by the stone. And what is the displacement of the stone​

Answer 1

Answer:

1 ) Let u be the initial velocity and h be the maximum height attained by the stone.

v

1

2

=u

2

−2gh,

(10)

2

=u

2

−2×10×

2

h

100=u

2

−10h ....(i)

Again at height h,

v

2

2

=u

2

−2gh

(0)

2

=u

2

−2×10×h

u

2

=20h ... (ii)

So, from Eqs. (i) and (ii) we have

100=10h

h=10m

Answer 2

Given: A stone is thrown vertically upwards. It travels to height 6 m and falls back to the ground.

To find: 1-distance and displacement when the stone reaches at topmost position

2- distance and displacement when the stone falls back in the hand

Explanation: When stone reaches the highest point, the velocity is zero. In the upwards journey, the height reached is 6 m. The distance is also 6m.

Since the displacement is the shortest distance between initial point and final point, it is also 6m.

When stone falls back in the hand, the distance travelled is the sum of distance covered in upward and downward journey, that is 6+6 = 12 m.

The displacement of the ball when it falls back is zero since there is no distance between starting and final point.

Therefore, the distance and displacement is 6 m in upward journey. When it falls back, distance and displacement is 12 m and 0 respectively.