Question

Suppose the kick is attempted 36.0m
from the goalposts, whose crossbar is 3.05 m above the
ground. If the football is directed perfectly between the
goalposts, will it pass over the bar and be a field goal?
Show why or why not. If not, from what horizontal distance
must this kick be made if it is to score?

apply x=xo+voxt+1/2axt^2
x0=0, x=36.0m, vox=16.0m/s, ax=0m/s^2
solver for t

apply y=y0+voyt+1/2ayt^2
y0=0, y=?, voy=12.0m/s, ay=-9.8m/s^2, t= previouslyfound
solve for y
i need step by step help please

Answer 1

you have a quadratic in t .... hard to read the notation tho

Answer 2

and they give you everything needed to substitute in to find y

Answer 3

yes it is sorry if its hard to read

Answer 4

how do we solve a quadratic equation? there are multiple methods available to us

Answer 5

you need to find the times that y = 3.05 m and plug the later one into x(t) to see if it is greater than 36 m

Answer 6

i can try and tell you what i geto for it . when id id itth e first time the numbers didn't look right

Answer 7

x=xo+voxt+1/2axt^2

x0=0, x=36.0m, vox=16.0m/s, ax=0m/s^2

36=0+16t+1/2(0)t^2

hmm, if ax=0 then t is linear

Answer 8

for x i got 2.2

Answer 9

so we are given a linear equation and asked to solve for the time, the time is then applied to an equation that models time spent in the air

36 = 16t, t = 36/16 = 18/8 = 9/4

Answer 10

solve for t with the y equation... you will get 2 solutions: one as the ball goes up and one as the ball comes down. you use the time for when the ball is coming down to see how far it has travelled.

Answer 11

so it takes 2.25 seconds to travel 36 meters, y is equal to or greater then the height of the goal post, then we have scored

Answer 12

y=y0+voyt+1/2ayt^2

y0=0, y=?, voy=12.0m/s, ay=-9.8m/s^2, t= previouslyfound

solve for y

y=0 +12t+1/2(-9.8)t^2

and t is known

Answer 13

\[3.05 = 0 +12t \frac{m}{\sec} -\frac{1}{2}\cdot 9.8t^2 \frac{m}{\sec^2}\]solve for t.

Answer 14

pgs approach will work, but its not 'according' to the problems instructions :)

Answer 15

7.1

Answer 16

thet're doing it but in a different way...

you can solve for t using the x equation and then plug in that time to the y equation and see if the height is above or below the goalpost.

Answer 17

\[36=0+16t \Rightarrow t=\frac{36}{16}=\frac{9}{4}\]

so

\[y \left( \frac{9}{4} \right)=0+12\left( \frac{9}{4} \right)-4.9\left( \frac{9}{4} \right)^2 = 2.19375 \text{ m}\]

Answer 18

no good

Answer 19

okay hold on ... i'mg ettingth at t=2.11

Answer 20

to find out how far, solve for t using the y equation with y(t) = 3.05m. plug that into the x equation and solve for x to find out how far to get a field goal with those initial conditions.

Answer 21

okay i will try and see

Answer 22

30.5=0+16.0m/s+2.19375+1/2+0t^2

Answer 23

okay so .163?

Answer 24

?
which part are you talking about?

Answer 25

or 24.81 the part where youte ll me to plug it backi n

Answer 26

if you solve for t using the y equation you get...
\[3.05 = 0 +12t - 4.9t^2 \Rightarrow t = 0.28804 \text{ & } 2.16093\]

\[x \left( 2.16093 \right)=16(2.16093) = 34.57488 \text{ m}\]

Answer 27

okay let me retype it and see...

Answer 28

okay yes i do get that number

Answer 29

now do we subtract 34.57488 and 2.19375m=32.38m

Answer 30

no! 34.57488 m is how far out in order to make the field goal.

the 2.19375 m is the height of the ball when kicked from 36 m out.

Answer 31

oh