Question

name:
54 the trajectory of a potato launched from a potato cannon on the ground at an angle of 45 degrees with an initial speed of 65 meters per second can be modeled by the parabola ( h(x) = x - 0.0023x^2 ), where the x - axis is the ground. find the height of the highest point of the trajectory and the horizontal distance the potato travels before hitting the ground.
there is a graph of the parabola here, with x - axis labeled horizontal distance (m) and y - axis labeled height (m)
a height: 109 m; distance: 435 m
b height: 121 m; distance: 418 m
c height: 118 m; distance: 421 m
d height: 102 m; distance: 409 m

Explanation:

Step1: Analyze the parabola function

The trajectory is modeled by \( h(x) = x - 0.0023x^2 \), which is a quadratic function in the form \( y = ax^2 + bx + c \) (here \( a=-0.0023 \), \( b = 1 \), \( c = 0 \)). For a parabola \( y=ax^2+bx + c \), the x - coordinate of the vertex (highest point for \( a<0 \)) is given by \( x=-\frac{b}{2a} \).

Substitute \( a=-0.0023 \) and \( b = 1 \) into the formula: \( x=-\frac{1}{2\times(-0.0023)}=\frac{1}{0.0046}\approx217.39 \)

Step2: Find the height at the vertex

Substitute \( x\approx217.39 \) into \( h(x)=x - 0.0023x^2 \)

\( h(217.39)=217.39-0.0023\times(217.39)^2 \)

First, calculate \( (217.39)^2\approx217.39\times217.39 = 47260.41 \)

Then, \( 0.0023\times47260.41\approx108.699 \)

\( h(217.39)=217.39 - 108.699\approx108.69\approx109 \)

Step3: Find the horizontal distance when hitting the ground

When the potato hits the ground, \( h(x) = 0 \), so we solve the equation \( x-0.0023x^2=0 \)

Factor out \( x \): \( x(1 - 0.0023x)=0 \)

This gives two solutions: \( x = 0 \) (launch point) and \( 1-0.0023x=0\Rightarrow x=\frac{1}{0.0023}\approx434.78\approx435 \)

Answer:

a. height: 109 m; distance: 435 m