A basketball is tossed from the top of a 3-mwall. The path of the basketball is definedby the relation y x2 2x 3, wherex represents the horizontal distancetravelled, in metres, and y represents theheight, in metres, above the ground. Howfar has the basketball travelled horizontallywhen it lands on the ground?
Question
A basketball is tossed from the top of a 3-mwall. The path of the basketball is definedby the relation y x2 2x 3, wherex represents the horizontal distancetravelled, in metres, and y represents theheight, in metres, above the ground. Howfar has the basketball travelled horizontallywhen it lands on the ground?
Solution
The basketball lands on the ground when y = 0. So, we need to solve the equation 0 = x^2 - 2x + 3 for x.
This is a quadratic equation, and we can solve it using the quadratic formula: x = [-b ± sqrt(b^2 - 4ac)] / 2a
In this case, a = 1, b = -2, and c = 3.
So, x = [2 ± sqrt((-2)^2 - 413)] / 2*1 = [2 ± sqrt(4 - 12)] / 2 = [2 ± sqrt(-8)] / 2
Since the square root of a negative number is not a real number, there are no real solutions for x.
This means that the basketball does not land on the ground. The equation given does not represent a possible path for the basketball.
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