Answer:
[tex]\displaystyle y_m=3.65m[/tex]
Explanation:
Motion in The Plane
When an object is launched in free air with some angle respect to the horizontal, it describes a known parabolic path, comes to a maximum height and finally drops back to the ground level at a certain distance from the launching place.
The movement is split into two components: the horizontal component with constant speed and the vertical component with variable speed, modified by the acceleration of gravity. If we are given the values of [tex]v_o[/tex] and [tex]\theta\\[/tex] as the initial speed and angle, then we have
[tex]\displaystyle v_x=v_o\ cos\theta[/tex]
[tex]\displaystyle v_y=v_o\ sin\theta-gt[/tex]
[tex]\displaystyle x=v_o\ cos\theta\ t[/tex]
[tex]\displaystyle y=v_o\ sin\theta\ t -\frac{gt^2}{2}[/tex]
If we want to know the maximum height reached by the object, we find the value of t when [tex]v_y[/tex] becomes zero, because the object stops going up and starts going down
[tex]\displaystyle v_y=o\Rightarrow v_o\ sin\theta =gt[/tex]
Solving for t
[tex]\displaystyle t=\frac{v_o\ sin\theta }{g}[/tex]
Then we replace that value into y, to find the maximum height
[tex]\displaystyle y_m=v_o\ sin\theta \ \frac{v_o\ sin\theta }{g}-\frac{g}{2}\left (\frac{v_o\ sin\theta }{g}\right )^2[/tex]
Operating and simplifying
[tex]\displaystyle y_m=\frac{v_o^2\ sin^2\theta }{2g}[/tex]
We have
[tex]\displaystyle v_o=20\ m/s,\ \theta=25^o[/tex]
The maximum height is
[tex]\displaystyle y_m=\frac{(20)^2(sin25^o)^2}{2(9.8)}=\frac{71.44}{19.6}[/tex]
[tex]\displaystyle y_m=3.65m[/tex]
