Math

QuestionFind the max horizontal distance of a shot with max height 23.3 units at 30 feet, following F(x)=0.02x2+1.2x+5.3F(x)=-0.02x^2+1.2x+5.3.

Studdy Solution

STEP 1

Assumptions1. The maximum height of the shot is23.3 units. . The maximum height occurs30 feet from the point of release.
3. The trajectory of the shot is defined by the equation (x)=0.02x+1.x+5.3(x)=-0.02x^+1.x+5.3.
4. The initial speed and the angle of launch are unknown.
5. The number "50" mentioned by the user is not relevant to the problem as it is unclear what it refers to.
6. We are looking for the maximum horizontal distance of the shot to the nearest tenth of a foot.

STEP 2

The equation of the trajectory is a quadratic equation, which represents a parabola. The maximum height of the shot corresponds to the vertex of the parabola. The x-coordinate of the vertex of a parabola given by the equation (x)=ax2+bx+c(x) = ax^2 + bx + c is given by b2a-\frac{b}{2a}.

STEP 3

Plug in the values for aa and bb from the given equation to find the x-coordinate of the vertex.
xvertex=1.22×0.02x_{vertex} = -\frac{1.2}{2 \times -0.02}

STEP 4

Calculate the x-coordinate of the vertex.
xvertex=1.20.04=30x_{vertex} = -\frac{1.2}{-0.04} =30

STEP 5

The x-coordinate of the vertex is the same as the horizontal distance at which the maximum height occurs. This is given as30 feet. Since the parabola is symmetric, the maximum horizontal distance of the shot is twice the x-coordinate of the vertex.
Maximumhorizontaldistance=2×xvertexMaximum\, horizontal\, distance =2 \times x_{vertex}

STEP 6

Plug in the value for xvertexx_{vertex} to calculate the maximum horizontal distance.
Maximumhorizontaldistance=2×30Maximum\, horizontal\, distance =2 \times30

STEP 7

Calculate the maximum horizontal distance.
Maximumhorizontaldistance=2×30=60Maximum\, horizontal\, distance =2 \times30 =60The maximum horizontal distance of the shot is60 feet.

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