Question 1171163: 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/-16 = 7/8
Substitute h to t:
x= -16t²+14t+6
= -16(7/8)² + 14(7/8) + 6
= 145/16
x= 9meters
Answer by ikleyn(52754)   (Show Source):
You can put this solution on YOUR website! .
Your solution is INCORRECT.
Notice that h(t) is the vertical coordinate - it is the HEIGHT.
You CAN NOT equate it to horizontal distance - they are TOTALLY DIFFERENT items.
I just answered you in my previous post THAT YOUR PROBLEM IS POSED INCORRECTLY,
because the height of the ring is not given.
Listen the instructions of the experts.
Otherwise, for what reason did you come to the forum ?
To demonstrate your incompetence ? - - - Thanks, we do not need it . . .
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