Well, the easy way is to say the average speed of the ball on the awy up is (12 + 0) / 2 = 6 m/s
Now how high did it go?
(1/2) m v^2 = m g h
144/2 = 9.8 h
h = 7.35 meters up
so
6 t = 7.35
t = 1.22 seconds
part 2 is trivial, same speed down as up but opposite direction, -12 m/s
A ball is thrown vertically upward with a speed of +12.0 m/s.
How long does the ball take to hit the ground after it reaches its highest point?
What is its velocity when it returns to the level from which it started?
2 answers
I do not like the way you did this questrion
1 answer