Answer:
y = 3.16 m
Explanation:
Once thrown straight up, the only influence on the ball is gravity, which acts slowing down the ball.
In absence of gravity, the ball had moved forever with the same initial velocity, so the displacement at a time t had been as follows:
y = v₀*t
Due to the action of gravity, the actual displacement is less than this value, and can be calculated with the following equation:
[tex]y = v0*t -\frac{1}{2} *g*t^{2}[/tex]
Replacing by v₀ = 12. 0 m/s, t = 0.300 s, and g = 9.8 m/s², we can solve for the vertical displacement, as follows:
[tex]y = v0*t -\frac{1}{2} *g*t^{2} = 12 m/s*0.300 s -\frac{1}{2} *9.8m/s2*(0.300s)^{2}\\= 3.6 m - 0.44 m = 3.16 m[/tex]
So, the position (taking the origin coincident with the point where the ball leaves the hand), is just the vertical displacement we have just found.
⇒ y = 3.16 m
