Question

A projectile is thrown with a velocity vat an angle a with horizontal. When the projectile is at a height equal to half of the maximum height, the vertical component of the velocity of projectile is

1. v sin a * 3
2. vsin a / 3
3. vsin a / root 2

Answer 1

$$\begin{gathered} InitialVerticalcomponentofthevelocityu=v\sin a \\ maximumheightisgivenby: \\ {v}^2-u^2=-2gh \\ attopmostheightv=0 \\ 0-{\left(v\sin a\right)}^2=-2gh \\ h=\frac{{\left(v\sin a\right)}^2}{2g} \\ h\text{'}=\frac{h}{2} \\ h\text{'}=\frac{{\left(v\sin a\right)}^2}{4g} \\ verticalcompoenetofthevelocityatthispoint \\ v{\text{'}}^2-u^2=-2gh\text{'} \\ v\text{'}=\sqrt{{\left(v\sin a\right)}^2-2g\frac{{\left(v\sin a\right)}^2}{4g}} \\ v\text{'}=\frac{v\sin a}{\sqrt{2}} \end{gathered}$$