Gr 12 U Physics (HARD QUESTION) (1 Viewer)

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A football kicker on his own 34 yard line wants to kick the ball so that it lads on the 1 yard line of the opposing team - for those of you who don't know about football, this is a 65 yard kick - use yards for this calculation not metres. The kicker can kick the ball at 27 yards/sec/ The kicker wants to give his own team mates as much time as possible to get to the spot where the ball is going to land. At what angle should the kicker kick the ball in order to get maximum hang time(time of flight)?

my note [note: instead of m/s, use yards/sec.]
1. The problem statement, all variables and given/known data

Given :
Distance(horizontal) = 65 yards
Delta Distance(the end height) = 0 yards
Velocity(of kicker) = 27 yards/sec.
Acceleration = -9.8m/s^2

2. Relevant equations
I think... delta D = velocity(vertical) x time + ( .5 x acceleration)(time^2)

trig--cosine, sine, and tan.

cos(theta)=velocity(vertical) / 27 yards
cos(theta)=velocity(horizontal) / 27 yards


3. The attempt at a solution
ok I've tried thinking of a few ways to start this, but I don't know where to start, If it's okay with anyone, can you give me a start push xD just to put me in the right direction =P don't do the work for me though I want to figure this out too =P

ok i think I might have this =P ill check bac kin 5 minutes

nvm...

What I was thinking was since there can only be one time that if the ball was kicked at 27m/s, for 33 yards, that would mean there would only be 1 angle in which can meet those requirements...
 
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LowlyPion

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First of all you want to convert all your measurements to m and m/s or else start thinking in yrds/s2 - not my choice.

Basically you have 2 unknowns T and θ

You know what the component of Vo is - Vo*Cosθ. And you know the relationship between that and distance X = V*T.

Your T is also related to Vo*Sinθ by T/2 = Vo*Sinθ/g
 
First of all you want to convert all your measurements to m and m/s or else start thinking in yrds/s2 - not my choice.

Basically you have 2 unknowns T and θ

You know what the component of Vo is - Vo*Cosθ. And you know the relationship between that and distance X = V*T.

Your T is also related to Vo*Sinθ by T/2 = Vo*Sinθ/g
sorry what is Vo? (velocity of object?)

and.. the component for Vo (horizontal component?)(X) is VoCosθ which is 27cosθ and the vertical component (Y) is VoSinθ which is 27Sinθ... but I don't have the time yet so I can't use any of the equations... but im gonna try reading your mini solution again =P

and the Yards/m^2 etc he said just keep it as that, don't mind it or anything because he wants us to think football xD so just pretend it says m/s
 

LowlyPion

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Since I see they want you to use yds instead of meters then 1 m = 1.094 yrd would look like a reasonable conversion factor for converting your g in m/s2.
 
Since I see they want you to use yds instead of meters then 1 m = 1.094 yrd would look like a reasonable conversion factor for converting your g in m/s2.
and the Yards/m^2 etc he said just keep it as that, don't mind it or anything because he wants us to think football xD so just pretend it says m/s

uhmm where did T/2 = Vo*Sinθ/g come from? , from your previous previous post oh ok i think i see where it came frrom =P
 

LowlyPion

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sorry what is Vo? ...
and the Yards/m^2 etc he said just keep it as that, don't mind it or anything because he wants us to think football xD so just pretend it says m/s
Vo is as you figured out the initial velocity or V at t=0.

Personally I'd convert the 9.81m/s2 using the factor 1.094 as mentioned before.

But if he explicitly said to treat m and yrds as equal ... then go with that.
 
Vo is as you figured out the initial velocity or V at t=0.

Personally I'd convert the 9.81m/s2 using the factor 1.094 as mentioned before.
ok so, the T/2 = Vo*Sinθ/g came from... d=vit + 1/2at^2? with the stuff subbed in?
 

LowlyPion

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ok so, the T/2 = Vo*Sinθ/g came from... d=vit + 1/2at^2? with the stuff subbed in?
Not quite. You can't mix vertical and horizontal like that.

Velocity in the Horizontal is given by X = Vo*Cosθ*T
 
Not quite. You can't mix vertical and horizontal like that.
oh rite. can you explain to me where T/2 = Vo*Sinθ/g came from then? lol like the general equation

edit,, oh and either way, wouldn't that mean I would have 2 unknown variables in 1 equation? which means I can't find the solution lol xD im probably wrong though

ok So, i know my vertical displacement is going to be zero, assuming that the field thingy is straight across no hills or anything. so, that would mean I can SORT of calculate the vertical component, by using d=VoSinθ + 1/2(-9.8)t^2
 

LowlyPion

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oh rite. can you explain to me where T/2 = Vo*Sinθ/g came from then? lol like the general equation

edit,, oh and either way, wouldn't that mean I would have 2 unknown variables in 1 equation? which means I can't find the solution lol xD im probably wrong though
Your time to max height - i.e. when vertical v = o is going to be when t = Vo*Sinθ/g. But since the Time at max height is only 1/2 way there, then it's 1/2*T = Vo*Sinθ/g

As to the rest of it ... don't get too many worries ahead.
 
Your time to max height - i.e. when vertical v = o is going to be when t = Vo*Sinθ/g. But since the Time at max height is only 1/2 way there, then it's 1/2*T = Vo*Sinθ/g

As to the rest of it ... don't get too many worries ahead.
ok sorry uhmm, so, the max height is when the velocity of the object is at the very peak, which is 0m/s, and where did ===> t = Vo*Sinθ/g. <==== the /g come from? I now understand why it's 1/2 *T

OHH ARE YOU USING a = V / t ==> t = V/a?
 

LowlyPion

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Yes.
 
Vo*Sinθ is your vertical velocity.

If V = a*t ... then ... t = V/a
yah thanks =P I was confused which equation you were using! lol i just figured that out xD ,, ok and this is where I get stuck. lol
 
Well what happens when you solve for θ and eliminate T?
is it possible to solve for θ when you have 2 unknown variables? O_O
 

LowlyPion

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You have 2 equations. Vertical, horizontal
ok so, now, do I do the same thing but with horizontal? because then the horizontal one wouldn't work because acceleration doesn't apply to horizontal...

ORR

do you mean I should substitute that into an equation and eliminate T that way..

equation :

d = vit + 1/2at^2?

and with subbing in the vertical values only,,

ok yyeah i need some help again please =P
 
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uhmm So, I subbed the vertical component into the horizontal one, for the time to eliminate it, than I ended up with

-323.4 = 27cosθ54sinθ

my quesiton now is, how do I Isolate the θ? lol!

can I do this?

??????????????????????????

-323.4 = 27*54(cosθsinθ)
-323.4 = 1458(cosθsinθ)
cos-1(sin-1(-323.4/1458)) = θ

???????????????????????????

nope that won't work either...
 

LowlyPion

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uhmm So, I subbed the vertical component into the horizontal one, for the time to eliminate it, than I ended up with

-323.4 = 27cosθ54sinθ

my quesiton now is, how do I Isolate the θ? lol!
Well there is the identity that 2CosθSinθ = sin2θ that you might find useful.
 
Well what happens when you solve for θ and eliminate T?
ok well from here I sort of got lost... how would I solve for θ and eliminate T? like where do I sub it in to?
 

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