contestada

When Luis's dad threw a bounce pass to him, the angles formed by the
basketball's path were congruent. The ball landed 2/3 of the way between them before it
bounced back up. If Luis's dad released the ball 40 inches above the floor, at what
height did Luis catch the ball?

Respuesta :

msm555

Answer:

20 inches above the floor

Step-by-step explanation:

Let's denote the height at which Luis caught the ball as [tex]h[/tex].

Given:

  • Luis's dad released the ball 40 inches above the floor.
  • The ball landed 2/3 of the way between the initial release point and the height [tex]h[/tex].

Since the ball is released 40 inches above the floor, the total distance traveled by the ball vertically is [tex]40 + h[/tex] inches.

According to the problem, the ball lands 2/3 of the way between the release point and [tex]h[/tex].

Therefore, the ball descends [tex]2/3[/tex] of the total distance:

[tex] \dfrac{2}{3}(40 + h) [/tex]

Now, Luis catches the ball just as it starts to ascend again. At this point, the ball has descended [tex]2/3[/tex] of the total distance and must ascend [tex]1/3[/tex] of the total distance.

Thus, Luis catches the ball at a height of:

[tex] h = \dfrac{1}{3}(40 + h) [/tex]

Now, solve for [tex]h[/tex]:

[tex] h = \dfrac{1}{3}(40 + h) [/tex]

[tex] 3h = 40 + h [/tex]

[tex] 2h = 40 [/tex]

[tex] h = 20 [/tex]

Therefore, Luis catches the ball at a height of 20 inches above the floor.