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# M1 - Projectiles question watch

1. (Original post by Daniel__)
After which you do what exactly?
rearrange to get u=
2. At somewhat of a loss at this point, anyone mind giving me a hint?
3. (Original post by Daniel__)
To 16tan35 - 4.9(16/u2cos235) - 9 = 0 ?
So, 4.9(16/u2cos235)= 16tan35-9

then

Work out the RHS, flip it and take the square root.
4. 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.
5. You have

Divide by a

Divide by b

Multiply by d

and you have 1/c

reciprocal to get c =
6. (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.
Note that I did it in two stages.

Firstly to get

Then I multiplied each side by and divided each side by
7. Then I multiplied each side by and divided each side by
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?
8. (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.
9. Strange tried that earlier and still am now getting a different answer.
10. (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.
11. (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
12. (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 in your calculation (correct) and it dropped to (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.
13. Aha thanks so much.
14. 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.
15. (Original post by Daniel__)
...
You have mis-understood the question

The horizontal range is 1250

So when the horizontal distance = 1250, then the vertical height = 0
16. Right thank you.

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Updated: March 9, 2013
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