A basketball is tossed from the top of a 3m wall. The path of the basketball is defined by the relation �y=-x^2+2x+3 where x represent the horizontal distance travelled, in metres, and y �represents the height, in metres, above the ground.

A) How far has the basketball travelled horizontally when it lands on the
ground?

3m. Because when I solved it, the roots were 0=*x+1)(x-3) and x-3=0 so x would be 3m.

B) What is the highest height the ball reaches?
Vertex is (1,4). Highest height would be 4m.

C) For how many seconds is the ball higher than 3.5 m?

How would I do this? Do I just plug 3.5 as y and solve? But that doesn't seem right.

a) there are two solutions, one is negative (ignore that).

0=-x^2+2x+3
0= x^2-2x-3=(x-3)(x+1)
so indeed, x=3
B) vertix is indeed 1,4
C)time in air. You need something else. It would be nice to know its horiztal velocity...

Ah, trying to solve math problems, huh? Well, let me give you a hand, or should I say, a clown nose! Here's how we can solve part C of your question.

To determine for how many seconds the ball is higher than 3.5m, we need to find the x-values where y is greater than 3.5. So, let's set up the equation:

- x^2 + 2x + 3 > 3.5

Now, let's simplify it a bit:

- x^2 + 2x - 0.5 > 0

Alright, time to use some clown magic! We can solve this quadratic inequality by finding the x-values where the parabola is above the x-axis. In other words, let's find the x-values where the equation is positive.

Using clown mathematics (also known as factoring), we can factor the quadratic inequality:

(x - 0.5)(x + 1) > 0

Now, we have two factors: x - 0.5 and x + 1. To find where the expression is positive, we can use the clown sign method, which involves considering the signs of the factors.

When x < -1, both factors are negative. When -1 < x < 0.5, x + 1 is positive, but x - 0.5 is negative. When x > 0.5, both factors are positive.

So, the clown conclusion is that the ball is higher than 3.5m for -1 < x < 0.5 seconds.

I hope this not-so-serious approach helped! Remember, math can be fun, even with a bit of clowning around!

To find the time for which the ball is higher than 3.5 m, we need to solve the equation y = 3.5, keeping in mind that y represents the height above the ground.

Substituting y = 3.5 into the equation, we get:

3.5 = -x^2 + 2x + 3

Rearranging the equation:

x^2 - 2x + 0.5 = 0

To solve this quadratic equation, we can use the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

In our case, a = 1, b = -2, and c = 0.5. Substituting these values into the quadratic formula:

x = (-(-2) ± √((-2)^2 - 4 * 1 * 0.5)) / (2 * 1)

Simplifying:

x = (2 ± √(4 - 2)) / 2

x = (2 ± √2) / 2

There are two potential values for x:

1) x = (2 + √2) / 2
2) x = (2 - √2) / 2

Since the basketball was tossed from the top of a 3 m wall, we are only interested in the positive value of x.

Thus, the basketball is higher than 3.5 m for the time it takes to travel (2 + √2) / 2 meters horizontally.

To find the time duration during which the ball is higher than 3.5 meters, you need to set up the given equation and solve it for the corresponding values of x.

The equation given is:
y = -x^2 + 2x + 3

To find the time when the ball is higher than 3.5 meters, you need to solve for x when y = 3.5. This is because x represents the horizontal distance traveled, and y represents the height above the ground.

So let's substitute y with 3.5 in the equation:
3.5 = -x^2 + 2x + 3

Now, you need to solve this quadratic equation to find the values of x. For simplicity, let's rearrange the equation into the standard quadratic form:
x^2 - 2x + 0.5 = 0

Now you can solve this quadratic equation for x using any preferred method, such as factoring, completing the square, or using the quadratic formula. Since this equation might not factor easily, we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)

For this equation, a = 1, b = -2, and c = 0.5. Let's substitute these values into the quadratic formula to find the solutions for x.

Applying the quadratic formula:
x = (-(-2) ± √((-2)^2 - 4(1)(0.5))) / (2(1))
x = (2 ± √(4 - 2)) / 2
x = (2 ± √2) / 2

Now, simplify the solutions:
x = (2 + √2) / 2 and x = (2 - √2) / 2

Since the height of the ball is greater than 3.5 meters, you are interested in the positive root. Therefore, x = (2 - √2) / 2.

Hence, the ball is higher than 3.5 meters for (2 - √2) / 2 seconds.