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Answer:
(8) The horizontal distance is 20 meters
(9) Maximum height is 2.5 meters
(c) The height when the horizontal distance is 7 meters is 2.275 meters
Step-by-step explanation:
Given
[tex]h = -0.025d^2 + 0.5d[/tex]
Solving (8): The horizontal distance which the ball lands
When the ball lands, the height is at 0.
So, we have:
[tex]h = -0.025d^2 + 0.5d[/tex]
[tex]0 = -0.025d^2 + 0.5d[/tex]
Rewrite as:
[tex]-0.025d^2 + 0.5d = 0[/tex]
Factorize
[tex]-0.025d(-d+ 20) = 0[/tex]
Split
[tex]-0.025d = 0\ or\ -d+ 20 = 0[/tex]
Solve for d
[tex]d = \frac{0}{-0.025}\ or\ -d = -20[/tex]
[tex]d = 0[/tex] or [tex]d = 20[/tex]
[tex]d = 0[/tex] ---- This represents the starting point
[tex]d = 20[/tex] ---- This represents the horizontal distance traveled
Solving (9): Maximum height
The maximum of a quadratic equation
[tex]y = ax^2 + bx + c[/tex]
is:
[tex]x = -\frac{b}{2a}[/tex]
So: [tex]h = -0.025d^2 + 0.5d[/tex] means that:
[tex]a = -0.025\ and\ b = 0.5[/tex]
The maximum is:
[tex]d = -\frac{b}{2a}[/tex]
[tex]d = -\frac{0.5}{2 * -0.025}[/tex]
[tex]d = \frac{0.5}{2 * 0.025}[/tex]
[tex]d = \frac{0.5}{0.05}[/tex]
[tex]d = 10[/tex]
Substitute [tex]d = 10[/tex] in [tex]h = -0.025d^2 + 0.5d[/tex] to calculate the maximum height
[tex]h = -0.025 * 10^2 + 0.5 * 10[/tex]
[tex]h = -2.5 + 5[/tex]
[tex]h = 2.5[/tex]
Solving (10): The height when the horizontal distance is 7m
Substitute [tex]d = 7[/tex] in [tex]h = -0.025d^2 + 0.5d[/tex]
[tex]h = -0.025 * 7^2 + 0.5 * 7[/tex]
[tex]h = -1.225 + 3.5[/tex]
[tex]h = 2.275[/tex]