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Answer 1

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; $\underline{y = \dfrac{111}{140} \cdot x - \dfrac{3}{140} \cdot x^2}$

• 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;

• $\underline{y = f(x) = \dfrac{111}{140} \cdot x - \dfrac{3}{140} \cdot x^2}$

b) The maximum height is given by the vertex of the parabola

The x-coordinate at the vertex is the point $-\dfrac{b}{2 \cdot a}$

Which gives;

$x-coordinate = \dfrac{\frac{111}{140} }{2 \times \dfrac{3}{140} } = 17.5$

The maximum height is therefore;

$f(x)_{max} = \dfrac{111}{140} \times 17.5 - \dfrac{3}{140} \cdot 17.5^2 \approx 7.334$

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

$Horizontal \ distance \ to \ maximum \ height = \dfrac{R}{2}$

Therefore;

$Horizontal \ distance \ to \ maximum \ height = \dfrac{37}{2} = 18.5$

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:

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