A ball is thrown to the sky directly from the edge of the skyscraper which has 61.9 meters lenght. Asssume that the ball does not hit the skyscraper when it returns from sky journey to the ground and hit the ground 4.0 seconds after it was thrown. Find the velocity of the ball when it was thrown (initial velocity)?

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.

powerful computers now, not then

their crystal patterns have been disrupted by stress