Question 2.SP.36: A ball is thrown at an angle of 40° to the horizontal. What ......

Choose the x and y axes with the origin at the point where the ball is thrown. Neglecting air resistance, the x component of the acceleration is zero. The y component of the acceleration is -g. With a_{x}=0 and a_{y}=-9.8 \mathrm{~m} / \mathrm{s}^{2}, Eq. (2.8) provides

d x=\left(v_{0}+a_{0} t\right) d t . \quad \int_{0}^{s} d x=\int_{0}^{t}\left(v_{0}+a_{0} t\right) d t . \quad \therefore s=v_{0} t+\frac{1}{2} a_{0} t^{2} \qquad (2.8)\\  \\x=v_{0 x} t \qquad \text { and } \quad y=v_{0 y} t-\frac{1}{2}(9.8) t^{2}

Given that when x=100, y=0, v_{0 x}=v_{0} \cos 40^{\circ} and v_{0 y}=v_{0} \sin 40^{\circ}, the above equations become

\begin{aligned}100 & =v_{0} t \cos 40^{\circ} \\0 & =v_{0} t \sin 40^{\circ}-\frac{1}{2}(9.8) t^{2}\end{aligned}

Solving the first equation for v_{0}, substituting this in the second equation, and solving for t give t=4.138 \mathrm{~s}. Substituting this value in the first equation yields v_{0}=31.5 \mathrm{~m} / \mathrm{s}. The maximum height occurs at one-half the distance. Therefore, for t = 2.069 s

y_{\max }=31.5 \times 2.069 \sin 40^{\circ}-\frac{1}{2} \times 9.8 \times 2.069^{2}=\underline{20.9 \mathrm{~m}}