The question goes: A rifle is fired with angle of elevation $36^{\circ}$. What is the muzzle speed if the maximum height of the bullet is 1600ft?
I ignored the x-component and cared only y-component.
position vector is $$r_y(t) = t(V_{\circ}\cos36^{\circ}-\frac{1}{2}gt^2)j$$
And I have to find a derivative form, which is a velocity $$r'_y(t)=V_{\circ}\cos36^{\circ}-gt$$ and set it equal to zero because at the maximum height the velocity is 0.
So all in all, $t=\frac{V_{\circ}}{g}\cos36^{\circ}$ and I insert this value into the position funtion, and comes out to be $$\frac{2(V_{\circ}\cos36^{\circ})^2-(V_{\circ}\cos36^{\circ})^2}{2g}=487.68m (1600ft)$$
And $V_{\circ}$ comes out to be $120.84m$ or $396ft$. But the answer in solution says to be $544ft$. Where did I do wrong?
$$r(y) = (\color{red}tV_{\circ}\color{red}\sin36^{\circ}-\frac{1}{2}gt^2)j$$
$$t=\frac{V_{\circ}}{g}\color{red}{\sin}36^{\circ}$$
$$\frac{2(V_{\circ}\color{red}\sin36^{\circ})^2-(V_{\circ}\color{red}\sin36^{\circ})^2}{2g}=487.68m (1600ft)$$
$$V_0 \approx \frac{\sqrt{487.68 (2g)}}{\sin 36^\circ}\times 3.28 \text{feet}$$