Let's write the law of motions of the ball on both x and y axis. The motion of the ball is a uniform motion on the x-axis, with constant velocity, and uniformly accelerated motion on the y-axis, with constant acceleration g:
[tex]x(t)= v_0 \cos \alpha t[/tex]
[tex]y(t)=v_0 \sin \alpha t - \frac{1}{2} gt^2[/tex]
where [tex]v_0 = 31 m/s[/tex] is the initial speed of the ball and [tex]\alpha=35 ^{\circ}[/tex] is the angle at which the ball is launched.
The ball reaches its maximum height when its vertical velocity is zero. The law for the vertical velocity is:
[tex]v_y(t)= v_0 \sin \alpha - gt[/tex]
By requiring [tex]v_y(t)=0[/tex], we find the time t at which the ball reaches the maximum height:
[tex]0=v_0 \sin \alpha -gt[/tex]
[tex]t= \frac{v_0 \sin \alpha}{g} = \frac{(31 m/s)(\sin 35^{\circ})}{9.81 m/s^2}=1.81 s [/tex]
And if we substitute this time t inside the law of motion on the y-axis, y(t), we find the maximum height of the ball:
[tex]h=v_0 \sin \alpha t - \frac{1}{2} gt^2=(31 m/s)(\sin 35^{\circ})(1.81 s)- \frac{1}{2}(9.81 m/s^2)(1.81s)^2= [/tex]
[tex]=16.11 m[/tex]
