Respuesta :
The correct answer is:
36 seconds; 9H.
Explanation:
This is a quadratic function in standard form. The value of b in standard form will be v₀, or the initial velocity.
When b changes and nothing else, H (the x-coordinate of the vertex) will change by the same percentage as b, but in the opposite direction. If the value of b doubles, the value of the x-coordinate of the vertex will also double. In this problem, we are tripling b, so the x-coordinate of the vertex will also triple.
This means that, since the vertex is the point where the line of symmetry goes through the graph, the root will also triple in value. This means instead of 12, it will be 12(3) = 36.
When the x-coordinate of the vertex is changed, the y-coordinate is changed by the square of the change to the x-coordinate. Thus when the x-coordinate is tripled, the y-coordinate will be changed by 3² = 9, which makes H now 9H.
The time taken by the slower stone is [tex]\boxed{4{\text{ sec}}}[/tex] and the maximum height is [tex]\boxed{9H}.[/tex]
Further explanation:
The formula for time can be expressed as follows,
[tex]\boxed{t = \dfrac{{v - u}}{g}}[/tex]
Given:
The speed of first stone is 3 times the initial speed of the other.
Explanation:
Part (a),
The final velocity of the stones is zero as they are thrown vertically upward.
Let us consider the speed of first stone is 3v.
The speed of the second stone is v.
The time taken by the first stone to reach the ground is [tex]{t_1}.[/tex]
[tex]{t_1} = \dfrac{{3v}}{g}[/tex]
The time taken by the second stone to reach the ground is [tex]{t_2}.[/tex]
[tex]{t_2} = \dfrac{{v}}{g}[/tex]
The ratio of [tex]{t_1}[/tex] and [tex]{t_2}[/tex] can be obtained as follows,
[tex]\begin{aligned}\dfrac{{{t_1}}}{{{t_2}}}&=\dfrac{{\dfrac{{3v}}{g}}}{{\dfrac{v}{g}}}\\\dfrac{{{t_1}}}{{{t_2}}}&=\dfrac{3}{1}\\\frac{{{t_1}}}{{{t_2}}}&=3\\\end{aligned}[/tex]
Substitute 12 for [tex]{t_1}[/tex] in above equation to obtain the value of [tex]{t_2}.[/tex]
[tex]\begin{aligned}\frac{{12}}{{{t_2}}}&= 3\\\frac{{12}}{3} &= {t_2}\\4&= {t_2}\\\end{aligned}[/tex]
Part (b)
The second equation of motion is [tex]H = ut + \dfrac{1}{2}g{t^2}.[/tex]
The maximumheight of slower stone can be expressed as follows,
[tex]H = \dfrac{{{v^2}}}{{2g}}[/tex]
The maximum height of faster stone can be expressed as follows,
[tex]\begin{aligned}x&= \frac{{{{\left( {3v} \right)}^2}}}{{2g}}\\&= \frac{{9{v^2}}}{{2g}}\\\end{aligned}[/tex]
The ratio of [tex]H[/tex] and [tex]x[/tex] can be obtained as follow,
[tex]\begin{aligned}\frac{H}{x} &= \dfrac{{\dfrac{{{v^2}}}{{2g}}}}{{\dfrac{{9{v^2}}}{{2g}}}}\\\frac{H}{x}&= \frac{1}{9}\\x&= 9H\\\end{aligned}[/tex]
The time taken by the slower stone is [tex]\boxed{4{\text{ sec}}}[/tex] and the maximum height is [tex]\boxed{9H}.[/tex]
Learn more:
- Learn more about inverse of the functionhttps://brainly.com/question/1632445.
- Learn more about equation of circle brainly.com/question/1506955.
- Learn more about range and domain of the function https://brainly.com/question/3412497
Answer details:
Grade: High School
Subject: Mathematics
Chapter: Speed
Keywords: Two stones, thrown vertically upward, speed, initial speed, 3 times, ground, faster stone, 12 second, return ground, slower stone, max height, H, faster stone, high.
