A ball is thrown straight up into the air, with an initial speed of 18.2 m/s. How far has it gone after 1.00s ?

Answer 1

First, find the time to the top of the trajectory, where v goes to zero. v = a*t, so t = v / a = 18.2 / 9.8.

This is greater than 1 second, so we know that the ball never gets to the top. Next, find the velocity after 1 second. The decrease is simply t*g, where g is the acceleration of gravity (9.8m/sec²)

So, the velocity after 1 second is 18.2 - 9.8 = 8.4 m/s

To get the distance, find the average velocity and multiply by 1 second.