A ball is thrown vertically upwards from a position p above the ground. It rises to the highest point q and returns to the point p. What is the net distance and displacement travelled by the ball
Answers
[ Refer the attachment ]
A ball is thrown from a point 'p' to point 'q' from the ground. The ball rises to point 'q' and then returns back to point 'p'.
We have to find the net distance and displacement travelled by the ball.
Distance is defined as the total distance covered by the body. Whereas displacement is defined as the shortest path travelled between the initial and final points.
Let's take an example to understand this.
Assume that the ball is thrown upto a height of 5 m. Means the distance between p and q is 5m. Reaching at point 'q' the ball returns back to point 'p' which is the initial point of the ball.
So, the total distance covered by the ball is 10 m. (5m from p to q point and 5 m from q to p point). OR twice of the distance covered by the ball i.e. 10 m.
As the initial and final points of the balls are same i.e. 'p' point. So, the displacement of the ball is zero.
Back to question!!
Net distance covered by the ball is twice of pq i.e. 2pq.
And the net displacement of the ball is zero.
Refer to the given attachment
Solution:-
Suppose the ball goes 10m above.
It means the distance between p and q is 10m.
Then,
→ The net displacement (S) = Final position - Initial position
→ Net displacement = distance move from point p to q + distance move forward point q to p
→Net displacement = 10m - 10m = 0
The net displacement will be zero because the final and initial position is same.
Now,
Net distance travelled by ball = total length of path travelled
= distance move from point p to q + distance move forward point q to p
= 10m + 10m
=20m
