Do the angles say nice things to each other or do they add to ninety degrees? Complimentary or complementary?
It's not really the minimum, it's the smaller maximum.
Enough correcting the question; let's answer it.
We have [tex]v =50[/tex] meters/sec. We're only interested in the y component of velocity.
If the first has
[tex]v_1 = v \cos \theta[/tex]
then the second is
[tex]v_2 = v\cos(90^\circ - \theta) = v \sin \theta[/tex]
We get a maximum when the velocity is zero, so for the first
[tex]0 = v_1 - g t_1 = v \cos \theta - g[/tex]
[tex]t_1 = \dfrac{v_1}{g} = \dfrac{v \cos \theta}{g}[/tex]
corresponding to a maximum height
[tex]h_1 = v_1 t_1 - \frac 1 2 g t_1^2 = v \cos \theta \dfrac{v \cos \theta}{g} - \frac 1 2 g (v^2 \cos^2 \theta)/g^2 = \dfrac{v^2 \cos^2 \theta}{2g} [/tex]
So
[tex]h_2 = \dfrac{v_0^2 \sin^2 \theta}{2g}[/tex]
We're given
[tex]30 = h=h_1-h_2[/tex]
[tex]h = \dfrac{v^2 (\cos^2 \theta - \sin^2 \theta)}{2g}[/tex]
That's the double angle formula but we'll avoid being tempted.
[tex] 2hg = v_0^2 (2 \cos ^2 \theta - 1)[/tex]
[tex] 2hg / v_0^2 = 2 \cos ^2 \theta - 1[/tex]
[tex] \cos ^2 \theta = \frac 1 2 + hg/ v_0^2[/tex]
[tex]h_1 = \dfrac{v_0^2 \cos^2 \theta}{2g} = v_0^2/4g + h/2[/tex]
[tex]h_1 = 50^2/40 + 15 = 77.5[/tex]
Choice 2
