Alignment of charges at the surface of an object producing an individual charge:
1 answer:
You might be interested in
Answer:
Explanation:
Yes , the horizontal distance travelled by the ball will depend upon the height of release .
When a ball is thrown at some angle from a height , it has two components , the vertical component and horizontal component . The ball goes in horizontal direction due to its horizontal component . Its vertical component has no role to play . But the horizontal range covered by the body thrown
depends upon the duration of time in which it remains in air . The longer it remains in air , the greater distance it can cover horizontally .
Horizontal distance covered = t x horizontal velocity
If V be the velocity of throw and Vx be its horizontal component
Horizontal distance covered = t x Vx
Now t depends upon the height . If height rises , time of fall will increase so horizontal distance covered will increase .
If h be the height from which the body is thrown , Vy be the vertical upward component of initial velocity
from the relation
s = ut + 1/2 at²
h = - Vy t + 1/2 at²
As h increases , t will increase and therefore horizontal distance covered will increase. If the ball has only horizontal velocity initially , Vy = 0
h = 1/2 gt²
Horizontal distance covered = t x Vx
=
From this expression also
Horizontal distance covered is proportional to .