This question is about equations reducible to quadratics:
Gerald hits his golf ball down the fairway. The path the ball followed was parabolic in shape with the relationship:
h=2d-1/30d^2
where h is the height above the ground of the golf ball when it is a horizontal distance d from him. Both h and d are in metres. How far from Gerald did the golf ball land?
0 = 2d -(1/30) d^2 = d [2 - (1/30)d ]
it is at ground level when d = 0 and when
2 - d/30 = 0
or when d = 60
To determine the horizontal distance at which the golf ball lands, we need to find the value of 'd' when the height 'h' becomes zero.
Given the relationship between height and horizontal distance as: h = 2d - (1/30)d^2
Setting h to zero and solving for 'd':
0 = 2d - (1/30)d^2
To simplify the equation, multiply both sides by 30 to get rid of the fraction:
0 = 60d - d^2
Rearranging the terms in descending order:
d^2 - 60d = 0
Now, we have a quadratic equation in the form of ax^2 + bx + c = 0, where a = 1, b = -60, and c = 0.
To solve this quadratic equation, we have a few options:
1. Factoring the equation:
In this case, we can factor out a common 'd':
d(d - 60) = 0
Therefore, two solutions are possible:
d = 0 or d - 60 = 0
However, the distance cannot be zero in this context since it represents the distance from Gerald. Hence, we discard the solution d = 0.
2. Using the quadratic formula:
The quadratic formula states that for any equation in the form ax^2 + bx + c = 0, the solutions for 'x' are given by:
x = (-b ± √(b^2 - 4ac)) / 2a
Applying this formula to our equation, we have:
d = (-(-60) ± √((-60)^2 - 4(1)(0))) / (2(1))
d = (60 ± √(3600)) / 2
d = (60 ± 60) / 2
Taking both the positive and negative values:
d1 = (60 + 60) / 2 = 120 / 2 = 60
d2 = (60 - 60) / 2 = 0 / 2 = 0
Again, we discard the distance of 0 as it does not make sense in this context.
Therefore, the golf ball lands at a distance of 60 meters from Gerald.