SOLUTION: 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+2

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?
Found 2 solutions by ikleyn, Clanther:
Answer by ikleyn(52778)   (Show Source): You can put this solution on YOUR website!
.

In order for the problem could be solved, THE HEIGHT of the ring should be given.

Without it, the problem CAN NOT be solved.

Your post does not provide the height of the ring.

Therefore, the problem, as it is worded, printed, posted and presented, is a FAKE.

You may post your "THANKS" to me for pointing your error.

I am open to accept your "THANKS".

============

To the @Clanther:

            Your solution is INCORRECT.

See my post under this link

https://www.algebra.com/algebra/homework/quadratic/Quadr

Answer by Clanther(6)   (Show Source): You can put this solution on YOUR website!
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

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