Answer:
[tex]\displaystyle x_{max}=21.57\ m[/tex]
[tex]\displaystyle y_{max}=3.78\ m[/tex]
Explanation:
Motion in Two-Dimensions
When an object is launched with a certain angle [tex]\theta[/tex] above ground level with an initial velocity [tex]\vec v_o[/tex], it describes a curve called a parabola, defined by the force of gravity which will eventually make the object return to the ground. There are two overlapping motions, the horizontal motion, at a constant speed, and the vertical motion, at changing speed because the acceleration of gravity modifies it. The velocity [tex]\vec v_o[/tex] is split into two components x,y
[tex]\displaystyle v_{ox}=|vo|\ Cos\theta[/tex]
[tex]\displaystyle v_{oy}=|vo|\ Sin\theta[/tex]
The position of the object is also split into its components, assuming the object was launched from ground level
[tex]\displaystyle x=v_{ox}.t[/tex]
[tex]\displaystyle y=v_{oy}.t-\frac{g.t^2}{2}[/tex]
The maximum horizontal distance the object reaches (called range) is
[tex]\displaystyle x_{max}=\frac{2v_{ox}\ v_{oy}}{g}[/tex]
The maximum height is given by
[tex]\displaystyle y_{max}=\frac{v_{oy}^2}{2g}[/tex]
The question doesn't ask for anything in particular, but to guide you in the solution of your own problem, we'll compute [tex]X_{max}[/tex] and [tex]Y_{max}[/tex] for you. The data is
[tex]\displaystyle |vo|=15\ m/s\ ,\ \theta =35^o[/tex]
Let's compute the range
[tex]\displaystyle x_{max}=\frac{(2)(15)\ cos35^o\ (15)\ sen35^o}{9.8}[/tex]
[tex]\displaystyle x_{max}=\frac{211.43}{9.8}=21.57\ m[/tex]
Now for the maximum height
[tex]\displaystyle y_{max}=\frac{(15\ . \ Sin35^o)^2}{2(9.8)}[/tex]
[tex]\displaystyle y_{max}=\frac{74.0227}{19.6}[/tex]
[tex]\displaystyle y_{max}=3.78\ m[/tex]
