Question

the path of a basketball shot can be modelled by the equation h= -1.3(d-0.09)²+2.9

what is the maximum height reached by the ball?

what is the horizontal distance of the ball from the player when it reaches the ground?

how far above the ground is the ball when the player releases it?
please write a clear answer or dont answer at all thnx

Answer 1

Answer:

The maximum height is 2.9 units, the horizontal distance of the ball from the player is 1.584 units away. The ball is 2.8895 units above the ground when released.

Step-by-step explanation:

Max Height:

The quadratic equation is in vertex form, which gives us the highest point by looking at h= -1.3(d-0.09)²+2.9 which is the y value of the vertex (how high or low the vertex is)

Horizontal Distance:

We can find this by finding the x intercept of the parabola,

set the h to zero.

h=-1.3(0-0.09)^2+2.9

like this and solve for h

How far above the ground:

we can find this by getting the y intercept by setting the d to zero.

0=-1.3(x-0.09)^2+2.9

-2.9=-1.3(x-0.09)^2

divide -2.9 by -1.3, remove the square by square rooting both sides, bring -0.09 over to the left and you get x