(a) At its maximum height, the ball's vertical velocity is 0. Recall that
[tex]{v_y}^2-{v_{0y}}^2=2a_y\Delta y[/tex]
Then at the maximum height [tex]\Delta y=y_{\mathrm{max}}[/tex], we have
[tex]-\left(\left(46\,\dfrac{\mathrm m}{\mathrm s}\right)\sin36^\circ\right)^2=2\left(-9.8\,\dfrac{\mathrm m}{\mathrm s^2}\right)y_{\mathrm{max}}[/tex]
[tex]\implies y_{\mathrm{max}}=37\,\mathrm m[/tex]
(b) The time the ball spends in the air is twice the time it takes for the ball to reach its maximum height. The ball's vertical velocity is
[tex]v_y=v_{0y}+a_yt[/tex]
and at its maximum height, [tex]v_y=0[/tex] so that
[tex]0=\left(46\,\dfrac{\mathrm m}{\mathrm s}\right)\sin36^\circ+\left(-9.8\,\dfrac{\mathrm m}{\mathrm s}\right)t[/tex]
[tex]\implies t=2.8\,\mathrm s[/tex]
which would mean the ball spends a total of about 5.6 seconds in the air.
(c) The ball's horizontal position in the air is given by
[tex]x=v_{0x}t[/tex]
so that after 5.6 seconds, it will have traversed a displacement of
[tex]x=\left(46\,\dfrac{\mathrm m}{\mathrm s}\right)\cos36^\circ(5.6\,\mathrm s)[/tex]
[tex]\implies x=180\,\mathrm m[/tex]
