M2 exam question

Ok ive been told to do a couple of papers before i return to college on tuesday and ive got stuck on this question

5. A particle is projected from a point with speed u at an angle of elevation w above the
horizontal and moves freely under gravity. When it has moved a horizontal distance x,
its height above the point of projection is y.
(a) Show that
y= xtan w - (gx^2 / 2u^2)(1+(tan^2)w)
(5 marks)

A shot-putter puts a shot from a point A at a height of 2 m above horizontal ground. The
shot is projected at an angle of elevation of 45° with a speed of l4m/s. By modelling the shot as a particle moving freely under gravity,
(b) find, to 3 significant figures, the horizontal distance of the shot from A when the shot hits the ground,
(5 marks)

(c) find, to 2 significant figures, the time taken by the shot in moving from A to reach the ground.
(2 marks)

Please help oh dont worry about part a

Reply 1

Jsk

Ive subbed in all of the information i know in part two but i just cant seem to solve it....
rep available to the best help

Reply 2

bruceleej

Jsk

Ive subbed in all of the information i know in part two but i just cant seem to solve it....
rep available to the best help

What have you done so far, write it up, so I can see where you have gone wrong.

For the first question- Write an equation for the verical motion and then the horizontal motion and eliminate the time 't'.

For the Second one- Write an equation for the verical motion of the partical bearing in mind that the vertical displacement of the particle is -2m. Then write an equation for the horizonatal motion{no gravity involed}. Next equate the times for both for equations, since the time taken for the vertical motion is equal to the time taken for the horizontal and that should give you the horizontal distance.

EDIT- I see youve done part a

Reply 3

Sanjetti

Ok I would use the integration method,

so if a = (0, -g)
v= (14Cos45, -gt + 14Sin45)
s= (14tCos45, -0.5gt2 + 14tSin45)

But that can be really covoluted, but its the only method I'm used to sorry

Reply 4

bruceleej

Sanjetti

Ok I would use the integration method,

so if a = (0, -g)
v= (14Cos45, -gt + 14Sin45)
s= (14tCos45, -0.5gt2 + 14tSin45)

But that can be really covoluted, but its the only method I'm used to sorry

Errmm, I think you method isnt covered in M2- for edexcel anyway.

Reply 5

Jsk

bruceleej

What have you done so far, write it up, so I can see where you have gone wrong.

into the big equation i have subbed y=2 u=14 w=45 g=9.8

and eventually got the equation x^2 - 20x + 40 =0 then used the quadratic formula and got x = 27.9 which is wrong... its 21.8

edit: sorry ive found my mistake.... g=-9.8 which makes the equation -40 on the end

Also did some silly errors in calculating them.... i think im trying to go too fast

Reply 6

Jsk

with the equation x^2 -20x -40 i seem to get no roots!

Reply 7

bruceleej

Jsk

into the big equation i have subbed y=2* u=14 w=45 g=9.8

and eventually got the equation x^2 - 20x + 40 =0 (this is right because its in the answers) then used the quadratic formula and got x = 27.9 which is wrong... its 21.8

edit: sorry ive found my mistake.... g=-9.8** which makes the equation -40 on the end

Also did some silly errors in calculating them.... i think im trying to go too fast

*y=-2m , not 2, because its now 2m below the point of projection not above.
**g=+9.8 not -9.8, I think the equation derived in 1 has already taken that into account.

Reply 8

bruceleej

Jsk

with the equation x^2 -20x -40 i seem to get no roots!

Well, you should...try again.

b^2 - 4ac = 400+160= 560 (>0)

So it does have 2 distinct roots.

Reply 9

Jsk

ok so i have the equation that it has in the answers which is the one above and i cant solve it

Reply 10

Jsk

bruceleej

Well, you should...try again.

b^2 - 4ac = 400-160= 240 (>0)

OMG! im being such an idiot... im getting the numbers for a,b,c mixed up.

Ah ok sorry mate

surely its 400+160? -4x-40x1=160

Reply 13

bruceleej

Jsk

surely its 400+160? -4x-40x1=160

yea, thats right sorry. Im just really tired and lazy.

Reply 14

Jsk

anyway i have done it! thank you soooooo much. reping you bruceleej for your help!

Reply 15

bruceleej

Jsk

anyway i have done it! thank you soooooo much. reping you bruceleej for your help!

Thanks very much, glad I could help even though i made loads of mistakes.

Reply 16

Jsk

hi im stuck on another question from the same exam paper..... ill see if i can find a link..... its question 7..... maybe i can attach it

M2_Jun_02.pdf601.3KB

Reply 17

Operator

There's no immediate way to solve Edexcel M2 statics questions.

You have to (i) resolve all forces in the horizontal and vertical directions, and (ii) take moments about a convenient point.

From here, solve the simultaneous equations thus obtained for the variable you want to find.