If a ball is thrown up at an initial speed of 40. m/s, how many seconds does it take to reach the top of its path?

Time, t = 4.08 secs

Explanation:

Given the following data;

Initial velocity, U = 40m/s

To find the time, we would use the first equation of motion;

Where;

V is the final velocity. U is the initial velocity. a is the acceleration. t is the time measured in seconds.

Making time, t the subject of formula, we have;

We know that acceleration due to gravity, g is 9.8m/s².

a = g = - 9.8m/s² because the ball is thrown in the opposite direction.  

Also, the final velocity is equal to zero (0) because the ball reached its maximum height.

Substituting into the equation, we have;

Time, t = 4.08 secs

Therefore, it will take the ball 4.08 seconds to reach the top.

Is a  vector quantity  that refers to "the rate at which an object changes its position." imagine a person moving rapidly - one step forward and one step back - always returning to the original starting position. while this might result in a frenzy of activity, it would result in a zero velocity. because the person always returns to the original position, the motion would never result in a change in position. since velocity is defined as the rate at which the position changes, this motion results in zero velocity. if a person in motion wishes to maximize their velocity, then that person must make every effort to maximize the amount that they are displaced from their original position. every step must go into moving that person further from where he or she started. for certain, the person should never change directions and begin to return to the starting position.