A Ball is thrown from the origin of a coordinate system. The equation of its path is y=mx-{(e^2m)x^2/1000}, where m is positive and represents the slope of the path of the ball at the origin.
For what value of m will the ball strike at the greatest height on a verticle wall located 100 feet from the origin?
you want to evaluate y(100). That is the height of the ball when it hits the wall
h(m) = y(100) = 100m-{(e^2m)/10}
So, where is that a maximum?
h' = 100 - e^(2m)/5
Where is h' = 0?
To find the value of m for which the ball strikes the greatest height on the vertical wall, we need to maximize the equation y = mx - (e^2m)x^2/1000.
The given equation represents a parabolic path of the ball. The maximum height of the ball occurs at the vertex of the parabola. In order to find the vertex, we need to convert the equation to vertex form.
The equation y = mx - (e^2m)x^2/1000 can be rewritten as:
y = - (e^2m)x^2/1000 + mx
To put it in vertex form, we need to complete the square.
The equation - (e^2m)x^2/1000 + mx can be rewritten as:
- (e^2m/1000)(x^2 - (1000/m)x)
Now, complete the square inside the parentheses:
- (e^2m/1000)(x^2 - (1000/m)x + (1000/2m)^2) + (e^2m/1000)(1000/2m)^2
Simplifying further gives us:
- (e^2m/1000)(x - (1000/2m))^2 + (e^2m/1000)(1000/2m)^2
We can rewrite this equation by multiplying by -1000/(e^2m):
y = -1000/(e^2m)(x - (1000/2m))^2 + (1000^2)/(4e^2m^2)
Comparing this equation with the standard form of a parabola y = a(x - h)^2 + k, we can determine that the vertex of the parabola is at the point (h, k) = ((1000/2m), (1000^2)/(4e^2m^2)).
Since we are looking for the maximum height on the vertical wall, we need to find the value of m that maximizes the y-coordinate of the vertex.
To find the maximum value, we need to take the derivative of the y-coordinate with respect to m and set it to zero:
(dy/dm) = (d/dm) [(1000^2)/(4e^2m^2)]
= -[(1000^2)/(2e^2m^3)]
Setting this derivative equal to zero gives us:
-(1000^2)/(2e^2m^3) = 0
Simplifying further:
m^3 = (1000^2)/(2e^2)
Taking the cube root of both sides:
m = (1000^2/(2e^2))^(1/3)
Therefore, the value of m that will make the ball strike the greatest height on the vertical wall, located 100 feet from the origin, is given by:
m = (1000^2/(2e^2))^(1/3)