An object that is in motion as a projectile follows a path or trajectory of a parabola
The function and values are;
- a) The equation of the quadratic function is; [tex]\underline{y = \dfrac{111}{140} \cdot x - \dfrac{3}{140} \cdot x^2}[/tex]
- b) The maximum height of the ball is approximately 7.334 m
- c) Horizontal distance at maximum height 18.8 meters
Reason:
a) Known parameters are;
Let f(x) = a·x² + b·x + c represent the equation of the parabola modelling the path of the ball, we have;
Points on the path of the parabola = (0, 0), (35, 1.5), 37, 0)
Plugging the values gives;
0 = a·0² + b·0 + c
Therefore, c = 0
1.5 = 35²·a + 35·b
0 = 37²·a + 37·b
Solving gives;
a = -3/140, b = 111/140
The equation of the quadratic function is therefore;
- [tex]\underline{y = f(x) = \dfrac{111}{140} \cdot x - \dfrac{3}{140} \cdot x^2}[/tex]
b) The maximum height is given by the vertex of the parabola
The x-coordinate at the vertex is the point [tex]-\dfrac{b}{2 \cdot a}[/tex]
Which gives;
[tex]x-coordinate = \dfrac{\frac{111}{140} }{2 \times \dfrac{3}{140} } = 17.5[/tex]
The maximum height is therefore;
[tex]f(x)_{max} = \dfrac{111}{140} \times 17.5 - \dfrac{3}{140} \cdot 17.5^2 \approx 7.334[/tex]
The maximum height of the ball is approximately 7.334 m
c) The distance the ball has travelled to horizontally is given by half of the range, R as follows;
The range of the motion, R = 37 meters
[tex]Horizontal \ distance \ to \ maximum \ height = \dfrac{R}{2}[/tex]
Therefore;
[tex]Horizontal \ distance \ to \ maximum \ height = \dfrac{37}{2} = 18.5[/tex]
The distance the ball has travelled horizontally to reach the maximum height horizontally 18.5 meters
Learn more about the trajectory of a projectile here:
https://brainly.com/question/13646224
