FOOTBALL. During a field goal attempt, the function h(d)= -0.02d+ 0.9d models the height, h(d) meters, of a football in terms of the horizontal distance, d meters, from where the ball was kicked. Find the horizontal distance the ball travels until it first hits the ground.

Question

please,please I need your help.
much appreciated in advance.

Answer 1

Your fomula can be rewritten h = 0.88d and is obviously wrong. The football does not keep rising in a straight line.

A correct formula would have the form:

h = - g (d/Vxo)^2 + Vyo*(d/Vxo).

where Vxo is the horizontal velocity component (which remains constant) and Vyo is the initial vertical velocity component

I believe you should have written
h(d)= -0.02d^2+ 0.9d
This would correspond to a football kicked with a horizontal velocity component of 22.1 m/s and an initial vertical component of 19.9 m/s.

Anyway, to do the proplem, set
h(d)= -0.02d^2+ 0.9d = 0 and solve for d, and take the answer that is not zero. You will get 45 meters, which is about 50 yards. that is typical for a field goal.

Answer 2

THANK YOU!!

Answer 3

one mini problem: i haven't learned this yet.
and i don't own a graphing calculator..

Answer 4

i haven't learned "h = - g (d/Vxo)^2 + Vyo*(d/Vxo). " this yet. woops.

Answer 5

You don't need a graphing calculator to solve -0.02d^2+ 0.9d = 0

Divide both sides by d and use algebra

0.02 d = 0.9

one more step and you have d.

Answer 6

oh.alright. i think i get it!! thanks

Answer 7

Never mind worrying about the formula h = - g (d/Vxo)^2 + Vyo*(d/Vxo).

I was just trying to derive the correct formula using physics, since the formula you provided was wrong.

Answer 8

one more question. -0.02d^2+ 0.9d = 0
where does the "-" in -0.02^2+0.9=0 go?

Answer 9

The minus sign goes where it is. You dropped the d^2 and the d in your version. You can't do that.

Add 0.02d^2 to both sides and you get
0.9 d = 0.02d^2

You seem to be unfamiliar with the basic rules and notations of algebra, and I recommend some tutoring if that is the case.