throwing a softball

Question

Aida threw a softball that traveled along a path described by the parabola y= -x^2 + 10x, where y = the height of the softball in feet and x = the horizontal distance in feet that the ball has traveled. Where does the softball hit the ground? How can you tell? What is the highest point reached during the flight of the ball? How do you know? What is the ball's horizontal distance from Aida when it is 24 feet up in the air? Does this make sense? Explain.

Answer 1

y= -x^2 + 10x, where y is height in feet and is horizontal distance in feet

Where does the softball hit the ground? How can you tell?

Hitting the ground means height (y) is 0 (assuming ground is at 0 feet :-) ). So, set y=0 and solve for x.

0 = -x^2 + 10x

We can use the quadratic equation but in this case, it is easier to simplify the equation and solve. First, move x out:

0 = x(-x + 10)

This equation has two solution (as most quadratic equations).

x = 0

and

(-x+10) = 0, then x = 10

x=0 makes sense, this is the starting point. x=10 is where it hit the ground.

What is the highest point reached during the flight of the ball? How do you know?

The parabola is symmetric. So, if it started at 0 height at x=0 and hit the ground (y=0 again) at x=10. So, the mid point has to be the highest point. (0 + 10)/2 = 5. Substitute x=5 in y = -x^2+10x =>

y = -(5)^2 + 10*5 = -25+50 = 25.

What is the ball's horizontal distance from Aida when it is 24 feet up in the air? Does this make sense? Explain.

y = 24, what is x?

24 = -x^2 + 10x

get all the terms on one side (left in this case)

x^2 - 10x + 24 = 0

Solve with quadratic equation:

x = (-b +- sqrt(b^2-4ac))/(2a), where a = 1, b = -10, c = 24

x = (-(-10) +- sqrt((-10)^2-4*1*24))/(2*1)

x = (10 +- sqrt(100-96)) / 2

x = (10 +- 2) / 2 (note that sqrt(4) is 2)

x1 = 12/2 = 6

x2 = 8/2 = 4

At both 4 and 6 feet horizontal distance, the ball was at height of 24 feet. This makes sense since the ball first went up and then down so crossed 24 feet height twice.