if a ball is thrown into the air at a speed of 22 m/s what is its position after 2.8 seconds?

Answer 1

Assuming it is thrown straight up, the altitude y after t seconds is

y = 22t - (g/2)t^2

where g = 9.8 m/s^2 is the acceleration of gravity

I get 23.2 meters

Answer 2

Thank you so much!