Projectile Motion Question (1 Viewer)

[Osiris8732]

A Long jumper left the ground at an angle of 15 degrees and jumped a distance of 7m. Considering this result, calculate the velocity of the long jumper when leaving the ground.
This questions from the success one textbook, they're answer uses a formula ive never seen before.
Does anyone know how to do this using the standard formulas we get on the formulae sheet?

 

[JayWalker]

A Long jumper left the ground at an angle of 15 degrees and jumped a distance of 7m. Considering this result, calculate the velocity of the long jumper when leaving the ground.
This questions from the success one textbook, they're answer uses a formula ive never seen before.
Does anyone know how to do this using the standard formulas we get on the formulae sheet?

Well you can split that up into Horizontal and Vertical Components.. Draw a triangle..

E.G

A_y = -9.8 [assuming this long jumper is on earth...]

Use a triangle and trig to determine intital velocities for y and x..

Remember that u_x = v_x

.. I would help further but I've lost motivation for helping this community...

 

[Jezzabelle]

i dunno how to work that without knowing time ? otherwise
(change in) x=ut ... would work

 

[JayWalker]

You figure out the time,, duh...

Let V_y=0, hence use projectile motion equations to discover t and then you get max height....

Not that the question actually asked that.... The question [as he has it there] only asked for initial velocities...

*mumbles something:: ... .... unthankful.. .. ... ...... honestly who doesnt know how to d......*

 

[Jezzabelle]

dont spos the answer is 11.7ms^-1

 

[Osiris8732]

Im still confused...
The question doesn't give any velocities at all...
it tells u 15 degrees and the distance travelled (7m)
how do we figure out any velocities from that?
assuming Vy = 0, we still cant figure out the time as we dont know Uy....

Could someone please show me all their working out for this question?

 

[Jezzabelle]

was 11.7 ms^-1 correct? if it was i will post my working

 

[jumb]

I got v = -6.5

stupid maths.

Edit, whoops forgot a minus. I got v = 6.5m/s

mised a v, i actually got 16.6

 

[Jezzabelle]

I got v = -6.5

stupid maths.

Edit, whoops forgot a minus. I got v = 6.5m/s

how can the initial velocity be negative if he is jumping up ?

edit: aha you edited just as i posted
wat i did was make t the subject by rearranging the x and y equations and then did a simultaneous equation to work out the value of U (initial velocity) x=ut , 7=ucos15.t therefore t=7.2/u
and did a sim thing for y, V=u+at, 0=usin15+(-9.8)t so, t= usin15/-4.9

t= 7.2/u=usin15/-4.9

then cross multiply
...35.28=-u²sin15
....u²=35.28/sin15
.... u²= 136.3114526... (on calculator)
therefore u=11.67524957ms^-1

 

[Heinz]

11.7 m/s is correct.

The formula given in the sucess one book requires some ext 1 maths knowledge to understand/derive. Its a pretty shitty (albeit easy) way of answering it since it assumes that students know the formula for range which #1 is limited to people who do ext 1 maths and #2 isnt given in the formula sheet.

 

[jumb]

Yeah, 11.7 is right.

 

[JayWalker]

how can the initial velocity be negative if he is jumping up ?

*looks at your blonde hair*.. ohh that explains it,
Having a negative answer should smack you in the face and force you to answer the question "VEctor or Scalar?!?!"
As velocity is ofcourse a vector, it can have a -ve sign, which means its going away from +ve, as im assuming right is +ve, then -ve means left, although the answer was wrong..

KTHXBAI

 

[Jezzabelle]

11.7 m/s is correct.

The formula given in the sucess one book requires some ext 1 maths knowledge to understand/derive. Its a pretty shitty (albeit easy) way of answering it since it assumes that students know the formula for range which #1 is limited to people who do ext 1 maths and #2 isnt given in the formula sheet.

i do ext1 maths but i didnt use that knowledge, i just used the physics formulas, see my edited post above

 

[Jezzabelle]

*looks at your blonde hair*.. ohh that explains it,
Having a negative answer should smack you in the face and force you to answer the question "VEctor or Scalar?!?!"
As velocity is ofcourse a vector, it can have a -ve sign, which means its going away from +ve, as im assuming right is +ve, then -ve means left, although the answer was wrong..

KTHXBAI

at least i got it fucking right you arse

in projectiles... they are postive on the upward journey and negative as they fall back down. thats what i was implying. obviously initially he is jumping up into the air, thus postive. shit head.

 

[jumb]

Yay, i finally got it. My maths skills have gone to the shitter in the last few days.

Stupid simultaneous equations.

anyway, my working:

(Δy = 0, Δx = 7)

Δy = -(9.8xt^2)/2 + vtsin15
(9.8xt^2)/2 = vtsin15

t = (vsin15)/4.9

-> Δx
Δx = vtcos15
7 = vcos15*(vsin15)/4.9
7 = (v^2 * sin30)/9.8
v = 11.7

Sorry that its really short hand, but whatever.

 

[Jezzabelle]

Yay, i finally got it. My maths skills have gone to the shitter in the last few days.

Stupid simultaneous equations.

anyway, my working:

(Δy = 0, Δx = 7)

Δy = -(9.8xt^2)/2 + vtsin15
(9.8xt^2)/2 = vtsin15

t = (vsin15)/4.9

-> Δx
Δx = vtcos15
7 = vcos15*(vsin15)/4.9
7 = (v^2 * sin30)/9.8
v = 11.7

Sorry that its really short hand, but whatever.

wat i did was make t the subject by rearranging the x and y equations and then did a simultaneous equation to work out the value of U (initial velocity) x=ut , 7=ucos15.t therefore t=7.2/u
and did a sim thing for y, V=u+at, 0=usin15+(-9.8)t so, t= usin15/-4.9

t= 7.2/u=usin15/-4.9

then cross multiply
...35.28=-u2sin15
....u²=35.28/sin15
.... u²= 136.3114526... (on calculator)
therefore u=11.67524957ms-1

 

[Heinz]

i do ext1 maths but i didnt use that knowledge, i just used the physics formulas, see my edited post above

Nice, didnt think of doing it that way. Theres a small typo "7=ucos15.t therefore t=7.2/t".

 

[Jezzabelle]

Nice, didnt think of doing it that way. Theres a small typo "7=ucos15.t therefore t=7.2/t".

thanks. just fixed it. thanks for being nice and not a misogynist pig like sum ppl...

 

[jumb]

wat i did was make t the subject by rearranging the x and y equations and then did a simultaneous equation to work out the value of U

I first off did that with the Δx and Δy equations and getting V, but fucked it up and got 6.5. So I took it slowly and did it the above way.

Not that it matters, will we even get that in the HSC? Seems too mathy for the modern day sciences.

 

[Jase]

hmm... well. I used 3U derivation. Do you do 3U maths?

If not, then you can memorise some new formulae:

Range - R = [V^2 sin(2@)] / g

Max Height - H = [V^2 (sin@)^2 ]/ 2g

and er.. Time of Flight is T = 2 x the time taken to reach max height...