The speed u of the object at the height of (1/2) hmax is [tex]0.707[/tex] [tex]v_{i}[/tex].
Explanation:
At a height that is half of the maximum height, the object will have only half of its initial kinetic energy, the rest half is converted to potential energy at this point. So,
[tex]1 / 2 m v^{2} _{i} = m (v_{mid})^{2} }[/tex]
[tex]v_{mid} = \frac{v}{\sqrt{2} }[/tex]
[tex]v_{mid} =[/tex] [tex]0.707[/tex] [tex]v_{i}[/tex].
The speed u of the object at the height of (1/2) hmax is [tex]0.707[/tex] [tex]v_{i}[/tex].
The speed u of the object at the height of (1/2)hmax is 0.707[tex]v_{i}[/tex].
At a height that is half of the maximum height the object will have only half of its initial kinetic energy the rest half is converted to potential energy at this point. so
[tex]\frac{1}{2} mv_{i} ^{2} = m (v_{mid} )^{2} \\[/tex]
[tex]v_{mid} = \frac{v_{i} }{\sqrt{2} }[/tex]
[tex]v_{mid} = 0.707 v_{i}[/tex]
So, The speed u of the object at the height of (1/2) hmax is 0.707[tex]v_{i}[/tex].
For more details on kinetic energy follow the link:
https://brainly.com/question/999862
