a/b. The ball has velocity vector at time [tex]t[/tex]
[tex]\vec v=(v_x,v_y)=(v_0\cos63^\circ,v_0\sin63^\circ-gt)[/tex]
where [tex]v_0=16\dfrac{\rm m}{\rm s}[/tex] is the ball's initial speed and [tex]g=9.8\dfrac{\rm m}{\mathrm s^2}[/tex].
c. At its highest point, the ball has 0 vertical speed. This occurs when
[tex]v_0\sin63^\circ-gt=0\implies t=1.5\,\mathrm s[/tex]
d. Recall that
[tex]{v_y}^2-{v_{0y}}^2=-2g\Delta y[/tex]
so that at its highest point,
[tex]0^2-(v_0\sin63^\circ)^2=-2g\Delta y\implies\Delta y=10\,\mathrm m[/tex]
e. This is just twice the time it takes for the ball to reach its maximum height, [tex]t=2.9\,\mathrm s[/tex].
f. The ball's horizontal position after time [tex]t[/tex] is
[tex]v_0\cos63^\circ\,t[/tex]
so that after the time found in part (f), the ball has traveled
[tex]v_0\cos63^\circ(2.9\,\mathrm s)=11\,\mathrm m[/tex]
