Question 1171140
While playing basketball, you are attempting to make a 3-point shot. The height in meters of the ball thrown at an angle of 45 degrees is given by the quadratic function h=-16t²+20t+6, where t is the time in seconds after throwing. The ball's horizontal distance in meters from you is modeled by x=6t. Assuming the ball went inside the ring, what is the horizontal distance from you to the ring?

Regarding this problem, I don't know if my solution is correct. I've came out with the answer given in the module which is 9 meters.

Here is my solution:

By analyzing the problem, the height of the ball as I hold it horizontally is given as h=-16t²+20t+6 and the height of the ball from me as I raised it ready to make a 3-point shot horizontally assuming that it will go inside the ring is x=6t.

So the distance of the ball from the ground -16t²+20t+6 is subtracted from the distance of the ball as I raised it x=6t.

Let x be the horizontal distance from me to the ring.

f(x)=-16t²+20t+6-6t
    =-16t²+14t+6

Solve for h in vertex formula h=-d/2a
      a=-16, b=14, c=6
      h= -14/2(-16) = 7/16

Substitute h to t:
      x= -16t²+14t+6
      = -16(7/16)² + 14(7/16) + 6
      = 145/16
      x= 9meters