Answer: 1.22 m
Explanation:
The equation of motion in this situation is:
[tex]y=y_{o}+V_{oy}t-\frac{g}{2}t^{2}[/tex] (1)
Where:
[tex]y=0[/tex] is the final height of the ball
[tex]y_{o}=h[/tex] is the initial height of the ball
[tex]V_{oy}=V_{o}sin(0\°)=0[/tex] is the vertical component of the initial velocity (assuming the ball was thrown vertically and there is no horizontal velocity)
[tex]t=0.5 s[/tex] is the time at which the ball lands
[tex]g=9.8 m/s^{2}[/tex] is the acceleration due gravity
So, with these conditions the equation is rewritten as:
[tex]h=\frac{g}{2}t^{2}[/tex] (2)
[tex]h=\frac{9.8 m/s^{2}}{2}(0.5 s)^{2}[/tex] (3)
Finally:
[tex]h=1.22 m[/tex]
