The relationship between the vertical height and the horizontal distance of a ball thrown at 45 degree is given by the equation y = x - x^2/50, where y is the vertical height and x is the horizontal distance. (a) Find the maximum vertical height reached. (b) Find the horizontal distance travelled.
(a) Take the derivative and solve for x when y' is 0. Evaluate y at that value.
(b) Solve the equation for x when y is 0.
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Alternatively, note that a ballistic curve of that form (quadratic) is symmetric around the maxima/minima and use the solution of one to solve the other.
To find the maximum vertical height reached and the horizontal distance traveled, we need to use the equation y = x - x^2/50.
(a) To find the maximum vertical height reached, we need to find the vertex of the equation. The vertex of a quadratic equation in the form y = ax^2 + bx + c is given by the formula x = -b/2a.
In our equation y = x - x^2/50, a = -1/50 and b = 1. Plugging these values into the formula, we get x = -1/(2(-1/50)) = 25.
To find the corresponding y value, we substitute x = 25 into the equation y = x - x^2/50:
y = 25 - (25^2/50) = 25 - (625/50) = 25 - 12.5 = 12.5.
Therefore, the maximum vertical height reached is 12.5 units.
(b) To find the horizontal distance traveled, we need to find the x-intercepts of the equation because the ball would have traveled its horizontal distance when it hits the ground (y = 0).
We set y = 0 in the equation y = x - x^2/50:
0 = x - x^2/50.
Simplifying, we get x - x^2/50 = 0. Multiply both sides by 50 to get rid of the fraction:
50x - x^2 = 0.
Rearranging, we have x^2 - 50x = 0. Factoring out x, we get:
x(x - 50) = 0.
Setting each factor to zero, we have x = 0 and x - 50 = 0. Solving for x in each case, we have x = 0 and x = 50.
Therefore, the horizontal distance traveled is 50 units.
So, (a) The maximum vertical height reached is 12.5 units, and (b) the horizontal distance traveled is 50 units.