M1 - Projectiles question

Scroll to see replies

Reply 20

TenOfThem

Original post by Daniel__

After which you do what exactly?

rearrange to get u=

Reply 21

Daniel__

At somewhat of a loss at this point, anyone mind giving me a hint?

Reply 22

ghostwalker

Original post by Daniel__

To 16tan35 - 4.9(16/u²cos²35) - 9 = 0 ?

So, 4.9(16/u²cos²35)= 16tan35-9

then

1/u^2=\frac{\cos^235}{4.9\times 16}(16\tan 35-9)

Work out the RHS, flip it and take the square root.

Reply 23

Daniel__

Confused at several parts:

When t is subbed giving you
Does the Usin35(16/ucos35) , cancel down to 16Utan35 or 16tan35

And as for the 4.9(16/u²cos²35)= 16tan35-9, not quite sure how you rearranged to get

. Forgive me for my terrible mathematical skills. :frown:

Reply 24

TenOfThem

You have

a(\frac{b}{cd}) = e

Divide by a

Divide by b

Multiply by d

and you have 1/c

reciprocal to get c =

Reply 25

ghostwalker

Original post by Daniel__

Confused at several parts:

When t is subbed giving you
Does the Usin35(16/ucos35) , cancel down to 16Utan35 or 16tan35

You've got one u in the numerator and one in the denominator, hence they cancel and you're left with the latter.

And as for the 4.9(16/u²cos²35)= 16tan35-9, not quite sure how you rearranged to get

. Forgive me for my terrible mathematical skills. :frown:

Note that I did it in two stages.

Firstly to get

4.9(16/(u^2\cos^2 35))= 16\tan 35-9

Then I multiplied each side by

\cos^235

and divided each side by

4.9\times 16

Reply 26

Daniel__

Then I multiplied each side by

\cos^235

and divided each side by

4.9\times 16

After which you flip to get the reciprocal of 1/u², giving you 4.9*16/cos²35(16tan35-9)? After which you solve the RHS and square root it?

Reply 27

ghostwalker

Original post by Daniel__

After which you flip to get the reciprocal of 1/u², giving you 4.9*16/cos²35(16tan35-9)? After which you solve the RHS and square root it?

Yep, that will work.

Reply 28

Daniel__

Strange tried that earlier and still am now getting a different answer.

Reply 29

ghostwalker

Original post by Daniel__

Strange tried that earlier and still am now getting a different answer.

I presume the correct one now, as you've not said.

Reply 30

Daniel__

Original post by ghostwalker

I presume the correct one now, as you've not said.

Oh apologies, its not 29.13.

I basically did (4.9*(16)/cos235)16tan35-9 then sqrted the answer

getting 16.04

Reply 31

ghostwalker

Original post by Daniel__

Oh apologies, its not 29.13.

I basically did (4.9*(16)/cos235)16tan35-9 then sqrted the answer

getting 16.04

Just realised that at one stage you had

16^2

in your calculation (correct) and it dropped to

16

(incorrect). I didn't notice and had carried on with that.

Also 16tan35-9 should be in the denominator after you flipped it. Make sure your brackets are accurate.

Aha thanks so much.

A cannon ball fired at an angle of 10º has a range, on a horizontal plane, of 1.25km. Ignoring air resistance, find the speed of projection

(↑):s=1250m u=ucos10 a=0 t=t
(→):s=? u=usin10 a=-9.8 t=t

How would I go about starting this given that I'm not told the vertical displacement nor the vertical u. I assume eventually once I've resolved it in both directions I'll just use Pythagoras to find the speed.