Question

ball is thrown upwards and
가
goes to the height of tom and
comes down.
What is
the Det displacement
what the net Distance
os​

Answer 1

Correct Question :

A ball is thrown upwards which goes to the height of 100 m and  comes down. What is  the net displacement and the Distance  covered by the ball ?

$\rule{200}{2}$

Explanation :

Displacement : Displacement is the shortest distance covered by the body or object. Displacement can be zero when the final and initial point of the object is same.

Now, In this question ball goes upward and also come back down.

So, The displacement of the ball is zero.

Distance : Distance is the total path covered by the object. Distance is always greater than the displacement.

Now, in this question the ball come backs.

So,

⇒ Distance = 100 + 100

⇒ Distance = 200

⇒ Distance = 200 m

Answer 2

CORRECT QUESTION -

Ball is thrown upward with a velocity of 20m/s.Calculate the max. height attained,net displacement and total distance covered by the ball(g=10m/s)

ANSWER -

Dear User,

Initial velocity , u = 20m/s

Acceleration due to gravity,g = 10 m/s

At the highest point, final velocity i.e. v becomes 0

Using v2-u2= 2gh, we get

(0)2-(20)2 = 2 x (-10) x s ...(g is taken as negative because it opposes the vertical motion)

or, 400 = 20 h

or h = 20m

Therefore, maximum height attained is 20 m

Total distance covered by ball = 2 x 20 = 40 m

Net displacement = Upward displacement + Downward displacement

= 20 + (-20) = 0 m