Answer:
a1) 61 seconds, a2) 31813 meters, a3) 4592 meters, b) [tex]y = x\cdot \tan 30^{\textdegree} -\frac{1}{2}\cdot (9.8\,\frac{m}{s^{2}} )\cdot \frac{x^{2}}{[(600\,\frac{m}{s} )\cdot \cos 30^{\textdegree}]^{2}}[/tex]
Step-by-step explanation:
a1) The parametric equations are:
[tex]x = (600\,\frac{m}{s} \cdot \cos 30^{\textdegree})\cdot t[/tex]
[tex]y = (600\,\frac{m}{s}\cdot \sin 30^{\textdegree} )\cdot t - \frac{1}{2}\cdot (9.8\,\frac{m}{s^{2}} )\cdot t^{2}[/tex]
The horizontal distance occurs when [tex]y = 0[/tex]. Then:
[tex]-4.9\cdot t^{2} + 300\cdot t = 0[/tex]
By factoring the expression, roots can be found easily:
[tex]t \cdot (-4.9\cdot t + 300) = 0[/tex]
The time needed is:
[tex]t = 61.225\,s[/tex]
a2) The horizontal distance evaluated at given time is:
[tex]x = 519.615\cdot (61.225\,s)[/tex]
[tex]x = 31813.428\,m[/tex]
a3) The maximum height reached by occurs when vertical velocity equals to zero. Vertical velocity can be found by deriving the function for vertical position:
[tex]v_{y} = 600\,\frac{m}{s}\cdot \sin 30^{\textdegree}- (9.8\,\frac{m}{s^{2}} )\cdot t[/tex]
[tex]300 - 9.8\cdot t = 0[/tex]
The instant associated with maximum height is:
[tex]t = 30.612\,s[/tex]
The maximum height is:
[tex]y = -4.9\cdot (30.612\,s)^{2}+300\cdot (30.612\,s)[/tex]
[tex]y = 4591.836\,m[/tex]
b) The equation is:
[tex]y = x\cdot \tan 30^{\textdegree} -\frac{1}{2}\cdot (9.8\,\frac{m}{s^{2}} )\cdot \frac{x^{2}}{[(600\,\frac{m}{s} )\cdot \cos 30^{\textdegree}]^{2}}[/tex]
