Answer
a) See the answer below.
b) $\rm 11.5^\circ,\; 2.04 \;s$
Work Step by Step
a)
A ball is fired at an angle $\theta$ above the horizontal from the ground at an initial speed of 50 m/s. It traveled 100 m to the east then it hits the ground again. Find the time of the whole trip and the initial releasing angle. (ignore air resistance).
b)
From the first formula,
$$100=50\;\cos\theta \;t_1$$
Thus,
$$t_1=\dfrac{2}{\cos\theta}\tag 1$$
Plug that into the second given formula;
$$0=50\sin\theta t_1-4.9t_1^2$$
$$0=t_1(50\sin\theta -4.9t_1)$$
$$50\sin\theta =4.9t_1 $$
Plugging from (1);
$$50\sin\theta =4.9\; \dfrac{2}{\cos\theta} $$
$$\dfrac{9.8}{50}=\sin\theta \cos\theta $$
Multiplying both sides by 2;
$$\dfrac{9.8}{25}=\overbrace{2\sin\theta \cos\theta }^{\sin2\theta} $$
Thus,
$$2\theta=\sin^{-1}\left[\dfrac{9.8}{25}\right]=23.08^\circ $$
Thus,
$$\theta=\color{red}{\bf 11.5^\circ}$$
Thus,
$$t_1=\dfrac{2}{\cos11.5}=\color{red}{\bf2.04}\;\rm s$$
