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A ball is thrown from the top of a building. It reaches a maximum height of 45 feet after traveling 10 feet horizontally and lands on the ground after traveling 40 feet horizontally. The path of the ball can be modeled by a parabola, where y is the height (in feet) and x is the horizontal distance traveled (in feet). At what height was the ball thrown?



Answer :

The ball was thrown at height of 40 feet and it followed the path of a parabola .

A parabola is a pane curve which is symmetrical along its axis and the path of a parabola is defined as the locus of moving point which is equidistant from a fixed point and a particular straight line.

Let us consider the parabola to be of the form y = ax² + bx + c

Now we know that after travelling 40 feet , the ball drops ,

Therefore at x=40 y=0

or,  0 = 40² a + 40 b + c or, 1600 a + 40 b + c = 0

Again   -(b/2a) = 10 and (4ac-b²)/4a = 45

Now we will solve the equation to find the values of a ,b and c

-b = 20a or , b = - 20a

Therefore substituting the value of b in the third equation we get:

or, 4ac -(-20a)² = 45 × 4a

or, 4ac + 400 a² = 90 a

or, 2c + 200 a - 45 =0

Now solving the equation we get c = 40

Now we know that at x = 0 , y = c or y = 40 .

Hence the ball was thrown from a height of 40 feet:

To learn more about parabola visit:

https://brainly.com/question/21685473

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