Answer:
[tex]V_{0}= 22.5\frac{m}{s}[/tex] and Ymax=25.8m
Explanation:
Velocity at any time is given by [tex]V=V_{0}sin\theta -gt[/tex]. but when the arrow is on the top its velocity is zero and if it is launched upward the angle is 90°, so.
[tex]0=V_{0} sin90-gt[/tex]
[tex]V_{0}=gt=9.8\frac{m}{s^{2}}.2.3s=22.5\frac{m}{s}[/tex]
At the maximun height, position is given by [tex]Ymax=V_{0}sin\theta.t-\frac{1}{2}gt^{2}[/tex], replacing [tex]Ymax=22.5\frac{m}{s}x2.3s-\frac{1}{2}x9.8x(2.3)^{2}=25.8\frac{m}{s}[/tex]
