Answer:
The velocity of the ball is 3.52 m/s.
Explanation:
A projectile is any object that moves under the influence of gravity and momentum only. Examples are; a thrown ball, a fired bullet, a kicked ball, thrown javelin, etc.
Given that the ball was thrown vertically upward on the top of a skyscraper of height 61.9 m. So that the velocity can be determined by;
u =
Where: u is the velocity of the object, H is the height and g is the gravitational force on the object. Given that: H = 61.9 m and g = 10 m/, then;
u =
=
u = 3.5185
The velocity of the ball is 3.52 m/s.
Answer:
2.2 meters
Explanation:
Potential energy, PE created by a charge, q at a radius r from the charge source, Q, is expressed as:
is Coulomb's constant.
#The electric field, at radius r is expressed as:
From i and ii, we have:
#Substitute actual values in our equation:
Hence, the distance between the charge and the source of the electric field is 2.2 meters