Horizontal and Vertical distance

Question

Grady is flying a kite on 50 ft of string. Its vertical distance from his hand is 10 ft more than the horizontal distance from his hand. Assuming that the string is being held 5 ft above ground level, find its horizontal distance from Grady and its vertical distance from the ground.

Answer 1

Draw a graph: 

Let point A be the point that Grady is holding the string

Let point B be the point that the string is attached to the kite

Draw a vertical line from point A upward

Draw a horizontal line from point B toward the vertical line

Let point C be the intersection of the horizontal and the vertical lines

Triangle ABC is a right triangle since BC and AC are perpendicular at point C, then:

AB = 50 ft (given)

Let: BC = x ft, then: AC = x + 10 ft, now using the Pythagoras' Theorem we can write:

(x + 10)² + x² = (50)^{2 => expand:}

x² + 20x + 100 + x² = 2500 => simplify:

2x² + 20x - 2400 = 0 => divide by 2:

x² + 10x - 1200 = 0 => factor: find 2 numbers that multiply to -1200, add to +10: (40 & -30):

(x + 40)(x - 30) = 0 =>solve using: Zero Product Property : if ab = 0, then either a = 0 or b = 0 (or both):

x + 40 = 0 => x = -40 => reject since the length can not be negative

x - 30 = 0 => x = 30 ft

The horizontal distance from Grady = BC = x = 30 ft

The vertical distance from Grady's hand = AC = x + 10 = 40 ft

But the vertical distance to ground is = 40 + 5 = 45 ft => since the string is being held 5 ft above the ground

 

Note: you can use the quadratic formula to solve the quadratic equation as follow:

if : ax² + bx + c = 0 , then:

x = [-b ±√(b² - 4ac)]/2a

in this case we have:

x² + 10x - 1200 = 0 => a = 1 , b = 10 , c = -1200:

x = [-10±√(10² - 4 * 1 * (-1200))]/(2 * 1)

x = [-10±√(100 + 4800)]/2

x = [-10±√4900]/2

x = [-10±70]/2

x = [-10 - 70]/2 = -80/2 = -40 => discard the negative length

x = [-10 + 70]/2 = 60/2 = 30 ft

The same as above.